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ELEMENTARY   TREATISE 


A    L   O    E    B    E    A 


DESIGNED    A3 


FIRST  LESSONS  IN  THAT  SCIENCE. 


H.  N.  ROBINSON,   A.   M., 

AUTHOR  OF  Ay  UmrVTIRSITY  EBITION  OP  ALaEBRA — AN  ELEMENTART  TREATISS 

ON  NATURAL  PHILOSOPHY — A  WORK  ON  GEOMETRY,  CONTAINING  PLANE 

AND   SPHERICAL   TRIGONOMETRY;   ALSO,  AUTHOR  OF  A  TEXT  BOOK 

ON  ASTRONOMY,  AJfD  SEVERAL  OTHER  MATHEMATICAL  WORKS. 


NINTH     ST  AND  AE-.D^  ]^J?PTliO;N'.__i 


CINCINNATI: 
PUBLISHED  BY  JACOB   ERNST, 

No.  112   MAIN   STREET. 
1856. 


Entered  according  to  Act  of  Congress  in  the  year  1850,  by 

H.  N.  ROBINSON, 

In  the  Clerk's  Office  of  the  District  Conrt  of  tlie  Uniteo  States,  for  the 

District  of  Ohio. 

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tOUCATlON  DEPT 


'.,  ..  B£%»:BiiftTfeEf}  BT  JAMES  *  CO., 
CIMCmNATI. 


PREFACE. 


Every  teacher  is  desirous  of  having  as  few  textbooks  in  his  school  as 
is  consistent  with  efficient  and  sound  instruction,  and  in  accordance 
with  this  object,  great  efforts  have  been  made  by  several  authors  to  pro- 
duce a  work  on  Algebra  that  would  be  a  proper  text  book  for  all  grades  of 
pupils.  But  in  this  all  have  failed,  and  given  up  the  point  in  despair. 
The  student  of  adult  age,  and  possessing  a  passably  disciplined  mind, 
requires  a  different  book  from  the  mere  lad,  who  is  just  commencing  the 
science.  If  we  put  a  child's  book  into  the  hands  of  a  young  man,  he 
will,  probably,  become  displeased  with  the  book,  and  possibly  imbibe 
prejudice  and  distaste  for  the  science  itself ;  and  if  we  put  a  logical  and 
philosophical  work  into  the  hands  of  a  child,  he  is  sure  not  to  comprehend 
it,  however  well  and  fluently  he  may  be  made  to  repeat  the  contents  of  its 
pages.  But,  nevertheless,  as  Algebra  is  the  groundwork  of  all  the 
mathematical  sciences,  and  is  of  itself  a  system  of  pure  logic,  it  is 
important  that  it  should  be  commenced  at  an  early  age — eleven  or 
twelve,  or  if  otherwise  well  employed,  thirteen  or  fourteen  is  a  more 
suitable  age. 

It  is  a  prevalent  impression  that  Algebra  should  not  be  commenced 
until  the  pupil  has  acquired  a  good  knowledge  of  Arithmetic,  but  this  is  a 
great  error.  The  impression  would  be  well  founded,  provided  Arithme- 
tic was  the  most  elementary  science,  and  Algebra  was  founded  on  Arith- 
metic ;  but  the  reverse  is  the  fact — Algebra  is  elementary  Arithmetic, 
and  no  one  can  acquire  a  knowledge  of  Arithmetic  in  an  enlarged  and 
scientific  sense,  without  a  previous  knowledge  of  Algebra.  Beyond  nota- 
tion, numeration,  and  the  four  simple  rules.  Arithmetic  is  not  a  science, 
but  a  sequel  to  all  sciences,  it  is  numerical  computation  applied  to  any- 
thing and  to  everything.  Proportion,  as  a  science^  is  the  comparison  of 
magnitudes,  and  belongs,  properly  speaking,  to  Algebra  and  Geometry; 
and  the  rule  of  three,  in  Arithmetic,  is  but  little  more  than  some  of  its 
forms  of  application.  Problems  in  mensuration  are  very  properly  to  be 
found  in  books  called  Arithmetics,  but  mensuration  is  no  part  of  the 

9S4226 


4  PREFACE. 

■clenco  of  Arithmetic,  it  is  a  part  of  Geometry,  and  for  a  good  under- 
standing of  it,  geometrical  science  must  be  directly  consulted. 

So  it  is  with  many  other  parts  of  Arithmetic,  the  science  is  else- 
where ;  and  to  have  a  scientific  comprehension  of  many  parts  of  com- 
mon Arithmetic,  we  must  go  to  general  Arithmetic,  which  is  emphati- 
cally Algebra ;  and  in  preparing  this  work,  we  have  given  constant 
attention  to  this  branch  of  the  subject,  as  may  be  seen  in  our  treatment 
of  fractious,  proportion,  progression,  the  roots,  fellowship,  and  interest. 

All  these  subjects  can  be  better  illustrated  by  symbols  than  by  num- 
bers ;  for  numbers  apply  to  everything,  and,  of  course,  can  be  made  to 
show  no  particular  thing  ;  but  not  so  with  symbols,  at  every  step  the 
particular  elements  are  all  visible,  and  the  logic  and  the  reason  is  as 
distinct  in  every  part  of  an  operation  as  is  the  result.  For  these  reasons. 
Arithmetic  should  be  studied  by  symbols,  as  it  is  in  many  parts  of 
Europe  ;  many  of  their  books,  entitled  Arithmetics,  are  as  full  of  signs 
and  symbols  as  any  Algebra  that  ever  appeared. 

The  prominent  design  of  the  author  has  been  to  adapt  this  treatise  to 
the  wants  of  young  beginners  in  Algebra,  and  at  the  same  time  not  to 
produce  a  mere  childish  book,  but  one  more  dignified  and  permanent, 
and  to  secure  this  end,  he  has  kept  up  the  same  tone  and  spirit  as 
though  he  were  addressing  mature  and  disciplined  minds. 

Great  care  has  been  taken  in  the  selection  of  problems,  and  all  very 
severe  ones  have  been  excluded,  and  all  such  as  might  be  difficult 
when  detached  and  alone,  are  rendered  simple  and  easy  by  their  con- 
nection with  other  leading  problems  of  kindred  character. 

To  bring  out  the  original  thoughts  of  the  pupil  has  been  another 
object  which  he  designed  to  accomplish,  and  the  illustrations  are  given 
in  such  a  way  as  to  command  the  constant  attention  of  the  learner,  and 
if  he  learns  at  all,  it  will  be  naturally  and  easily,  and  what  he  learns 
will  become  a  part  of  himself. 

In  this  work,  great  importance  is  attached  to  equations,  not  merely 
in  solving  problems,  but  they  are  used  as  an  instrument  of  illustrating 
principles,  and  their  application  is  carried  further  in  this  book  than  in 
any  other  known  to  the  author. 

For  instance,  we  have  illustrated  the  nature  of  an  equation  by  the^ 
aid  of  simple  problems  in  subtraction  and  division  ;  and  conversely,  the 
simple  principle  of  equality  is  used  to  deduce  rules  for  subtraction, 
division,  the  reduction  of  fractions  to  a  common  denominator,  the  mul- 
tiplication of  quantities  afTected  by  different  fractional  exponents,  &-c. 

Notwithstanding  that  this  book  is  designed  to  be  practical,  it  contains 
more  illustrations,  and  is  more  theoretical  and  scientific  as  far  as  it  goes, 
than  any  other  book  desiigned  for  the  same  class  of  pupils. 


PREFACE.  5 

We  have  not  given  demonstrations  of  the  binomial  theorem,  nof 
made  any  investigations  of  logarithms,  or  the  higher  equations,  for  these 
subjects  belong  exclusively  to  the  higher  order  of  Algebra,  and  will  be 
found  very  clear  and  full  in  the  University  Edition  of  Algebra  by  the 
same  author. 

In  relation  to  great  generalities,  all  books  on  the  same  science,  are,  in 
substance,  much  alike,  yet,  in  the  clearness  and  distinctness  with  which 
they  present  principles,  they  may  be  very  different ;  and  to  arrive  at 
perfection  in  this  particular,  is,  and  should  be,  the  highest  ambition  of 
an  author. 

For  peculiarities  in  this  work,  the  teacher  is  respectfully  referred  to 
abbreviations  generally  in  solving  equations,  to  the  philosophical  uses 
made  of  equations  in  demonstrating  principles — the  formation  of  prob- 
lems, and  the  manner  of  arriving  at  arithmetical  rules,  which  may  be 
found  in  various  parts  of  this  work. 


Digitized  by  tine  Internet  Archive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementarytreatiOOrobirich 


CONTENTS 


Page 

Introduction 9 

Axioms 16 

Simple  problems  for  exercises 17 

Definition  of  terms 21 

SECTION    I. 

Addition 25 

Subtraction 31 

Subtraction  illustrated  by  Equations 35 

Multiplication 36 

The  product  of  minus  by  minus  illustrated 37 

Division   45 

Negative  exponents  explained 49 

Division  in  compound  quantities 51 

Factoring 55 

Multiple  and  least  common  multiple 58 

Algebraic  Fractions 62 

Complex  fractions 68 

Multiplication  of  fractions 70 

Division  of  fractions 73 

Division  illustrated  by  Equations 76 

Addition  of  fractions   78 

Addition  of  fractions  by  Equations 79 

Subtraction  of  fractions 85 

Subtraction  illustrated  by  Equations 86 

SECTION    II. 

Equations 89 

Transposition 93 

General  rule  for  reducing  equations 93 

Proportion,  as  applied  to  equations 98 


8  CONTENTS. 

Pagb. 

Questions  producing  simple  equations 101 

How  to  propose  convenient  problems 107 

How  particular  numerals  are  brought  into  problems Ill 

Equations  having  compound  fractions ' 116 

Equations  containing  two  unknown  quantities 120 

Three  methods  of  elimination 122 

Equations  containing  three  or  more  unknown  quantities — Rule 

for  elimination 130 

Questions  producing  equations  containing  three  or  more  un- 
known quantities 134 

Negative  results,  how  understood 137 

SECTION    III. 
Involution 140 

Expansion  of  a  binomial 144 

Application  of  the  binomial 148 

Evolution 150 

How  to  extract  roots  of  polynomials 153 

Approximate  rule  for  cube  root 1 63 

Product  of  quantities  affected  by  different  fractional  exponents — 
Art  83 167 

SECTION    IV. 

Equations  of  the  second  degree 171 

Pure  Equations 172 

Problems  producing  pure  equations 174 

Rules  for  completing  a  square 180 

Resolving  a  quadratic  expression  into  two  factors 186 

Questions  giving  rise  to  quadratic  equations 191 

Homogeneous  and  symmetrical  equations 193 

SECTION    V. 

Arithmetical  progression 204 

Examples  in  arithmetical  progression 211 

Geometrical  progression 213 

Examples  in  geometrical  progression 215 

General  problems  that  involve  progression 220 

Proportion,  theoretically  considered 224 

Method  of  deriving  certain  arithmetical  rules 236 

Fellowship,  theoretically  considered 237 


INTRODUCTION. 


Algebra  is  the  science  of  computation  by  mekiis  of  sym- 
bols. Letters  of  the  alphabet  are  generally  used  to  represent 
quantities  or  numbers,  and  conventional  signs  are  employed 
to  represent  operations,  and  to  abridge  and  generalize  the 
reasoning  in  relation  to  propositions  or  problems. 

We  sometimes  meet  with  persons  who  can  readily  solve 
quite  difficult  problems,  and  yet  are  not  able  to  explain 
the  steps  in  the  process  :  they  call  their  operations  working  in 
the  head — and,  indeed,  their  reasoning,  properly  written  out, 
is  Algebra;  but  not  having  a  knowledge  of  the  signs,  and 
possessing  no  skill  in  writing  out  the  thoughts  of  the  mind, 
they  do  not  know  that  it  is  Algebra. 

This  natural  adaptation  of  the  mind,  to  solve  problems  with- 
out the  aid  of  writing  down  the  operation,  is  very  essential  to 
success  in  this  science.  But  the  mind  can  only  go  a  very 
short  distance,  unaided  by  the  pen ;  nor  is  it  important  that 
it  should,  for  the  aid  given  by  that  instrument  is  efficient  and 
complete,  secures  the  ground  gone  over,  and  leaves  the  mind 
free  to  advance  indefinitely. 

In  a  purely  mental  process,  the  mind  must  retain  all  the 
results  thus  far  attained,  and  continue  the  reasoning  onward 
at  the  same  time.  And  this,  carried  to  excess,  breaks  down 
the  mind  rather  than  strengthens  it ;  and  for  this  reason,  a 
mere  mental  Algebra  must  be  regarded  as  one  of  the  ephe- 
meral efforts  of  the  times.  But  let  no  reader  construe  these 
sentiments  into  a  disapproval  of  mental  Algebra.  Every  Al- 
gebra, properly  understood,  is  mental  Algebra ;  for  the  mental 
process — the  reasoning  power — must  precede  every  operation. 


10  ELEMENTARY    ALGEBRA. 

To  compare  the  common  operations  of  the  mind  with  the 
brief  and  refined  language  of  science,  we  propose  the  follow- 
ing problems.  But  before  we  use  algebraical  language,  we 
must  explain  some  pi  ?ts:  §yiiibols,  and  here  we  insert  only 
those  intended  for  immediate  use. 

J.,  ..  I  "'  .',•■'-'  ■„  _     »   ,'.,       •  (fc 

,   :;:      .  -THE   SIGNS. 

1.  The  perpendicular  cross,  thus  +,  caWed  plus,  denotes 
addition. 

2.  The  horizontal  dash,  thus  — ,  called  minusy  denotes 
subtraction.*  These  signs  are  written  before  the  quantities 
to  which  they  are  aflfixed. 

3.  The  diamond  cross,  thus  X,  or  a  point  between  two 
quantities,  denotes  multiplication.  For  example,  5X  4,  or  5*4.. 
shows  that  4  and  5  must  be  multiplied  together. 

4.  A  horizontal  line  with  a  point  above  and  below,  thus  -^, 
denotes  division;  also  two  quantities,  one  above  another,  as 

numerator  and  denominator,  as  _  or  ^,  also  indicates  division, 

7       h 

and  shows  that  3  must  be  divided  by  7,  and  a  must  be  divided 

hyb. 

5.  Double  horizontal  lines,  thus  =,  represent  equality,  and 
show  that  the  quantities  between  which  it  is  placed  are  equal. 

6.  A  number  or  letter  before  any  quantity  shows  how  many 
times  the  quantity  is  taken,  and  is  called  the  coefficient  of  the 
quantity,  thus  3x,  shows  that  the  quantity  x  is  taken  3  times, 
and  nx  shows  that  the  quantity  represented  by  x  is  taken  as 
many  times  as  there  are  units  in  n. 

7.  A  vinculum  or  bar ,  or  parenthesis  (   ),  is  used 

*  The  signs  plus  and  minus,  in  general  science,  have  a  far  more  com- 
prehensive meaning  than  is  here  expressed.  Here  they  denote  simply 
what  is  to  be  done  with  the  quantities  to  which  they  are  attached  ;  but 
in  philosophical  problems,  they  may  denote  the  essential  value  of  the 
quantities,  as  credit  and  debt ;  and  in  geometry  tliey  may  represent  jjosi- 
tions,  as  north  and  sotrfA,  or  to  the  right  or  left  of  a  zero  line,  &c. 


INTRODUCTION. 


11 


to  connect  several  quantities  together.  Thus,  a4-6  or  {a-\-b) 
shows  that  a  and  b  are  there  to  be  considered  as  connected,  or 
making  but  one  quantity. 

We  now  turn  our  attention  to  the  problems — not  for  the 
purpose  of  finding  the  answers  to  them,  as  the  mere  arith- 
metical student  might  suppose — but  for  the  purpose  of  teach- 
ing the  manner  of  solving  them  by  the  scieiice  of  Algebra. 

1.  A  father  divided  120  cts.  among  his  three  sons.  Be  gave 
the  youngest  a  certain  number,  the  second  10  cerds  more,  and  the 
eldest  10  cents  more  than  the  second.  What  sum  did,  each 
receive  ? 


By  the  use  of  cmnmon  lan- 
guage^  this  question  may  be 
solved  thus : 

The  youngest  son  had  a 
share  of  the  money  ;  the  sec- 
ond son  had  a  like  share  and 
10  cents  more  ;  and  the  eldest 
had  also  a  like  share  as  the 
youngest,  and  20  cents  more. 
Therefore  the  three  boys  had 
3  shares  and  30  cents ;  but 
the  three  boys  had  120  cents, 
hence,  3  shares  and  30  cents 
are  equal  to  120  cents,  or  the 
three  shares  are  worth  90 
cents,  and  one  share  is  worth 
30  cents,  which  is  the  sum 
given  to  the  youngest. 


By  algebraical  language, 
thus: 

Let  X  represent  the  share 
of  the  youngest.  Then  by 
the  conditions 

a;=  3d  son's  share, 
a;-f  10=  2d     " 
ar-{-20=  1st    "        " 

3a;-f30=120  by  add. 

This  expression  is  called  an 
equation,  and  the  quantities 
on  each  side  of  the  sign  of 
equahty  are  called  members, 
or  sides  of  the  equation. 

It  is  an  axiom,  that  equals 
from  equals  the  remainders 
must  be  equal ;  and  in  this 
equation,  if  we  *take  30  from 
both  members,  we  have 

3a:=  120— 30=90 

Dividing  both  members  by 
3  gives  a;=30,  tlie  share  of 
tlie  youngest. 


1 


12 


ELEMENTARY    ALGEBRA. 


2.  If  75  dollars  he  added  to  a  share  in  a  certain  bridge  coni' 
pany,  the  sum  will  he  the  value  of  4  shares.  What  is  the 
value  of  a  share? 


BY     COMMON     LANGUAGE. 

One  share  and  75  dollars  is 
the  same  as  4  shares,  there- 
fore 75  dollars  is  3  shares, 
and  one  share  is  25  dollars, 
the  third  part  of  3  shares. 


ALGEBRAICALLY. 

Let  X  represent  the  value 
of  a  share. 

Then   x-^15=Ax. 

Taking  x  an  equal  quantity 
from  both  members  and 
75=3;r. 

Dividing  by  3  gives  25=ar. 

3.  A  gentleman  purchased  a  horse y  a  chaise  and  a  harness,  for 
$230.  The  harness  cost  a  certain  sum,  the  chaise  3  times  as 
much  as  the  harness,  and  the  horse  $20  more  than  the  chaise. 
Required  the  price  of  each. 

BY     COMMON     LANGUAGE. 

The  harness  cost  a  certain 
share  of  the  money,  and  the 
chaise  cost  3  such  shares,  and 
the  horse  cost  3  such  shares 
and  20  dollars  more ;  there- 
fore the  whole  cost  7  shares 
and  20  dollars,  which  must 
make  230  dollars.  Take  the 
20  dollai's  away,  and  the  7 
shares  is  the  same  as  210  dol- 
lars. Therefore  1  share  is  30 
dollars,  the  value  of  the  har- 
ness, and  $90  is  the  value  of 
the  chaise,  and  $1 10  the  value 
of  the  horse. 

4.  In  a  certain  school  ^  of  the  pupils  are  learning  geometry , 
\  are  learning  Latin,  and  10  more,  which  comprise  all  in  the 
school t  are  learning  to  read.     What  was  the  whole  number.^ 


ALGEBRAICALLY. 

Let  x=   the   value 

of  the  harness. 

Then  3x-   the 

of  the  chaise. 

And   3a;-[-20=    the 
of  the  horse. 


value 


value 


Sum 


Taking 


both 


7a;4-20=230. 
equals   from 
members  7a:=210 

Or  a;=30  by  division. 


INTRODUCTION.  13 


Br    COMMON     LANGUAGE. 

One-third    and    one-fourth 
added     together     make     ■^-^. 


ALGEBRAICALLY. 

Let    x=^    the    number    in 
school. 

Then  by  the  conditions 


X       X 

34-4+10=0;. 


Therefore  -^-^  is  the  number 
learning  to  read,  which,  by 
the  problem,  is  10;  hence  Jj 
of  the  number  in  the  school 
is  2,  and  the  whole  number  is 
24,  the  number  required. 

This  equation  may  be  troublesome  on  account  of  the  frac- 
tions ;  but  in  due  time  we  shall  give  rules  to  clear  equations 
of  fractions;  however,  fractions  here,  are  just  the  same  as 
fractions  elsewhere.  One-third  and  one-fourth  of  anyth'mg  is 
j\  of  that  thing;  therefore,  ^|-f-10=Y¥.  Now  from  the 
two  equals  take  ^f,  and  10=ff ;  dividing  by  5,  ^—j^\  mul- 
tiplying by  12  gives  24=a?,  the  final  result. 

From  these  examples  it  will  be  perceived  that  Algebra  is 
but  an  artificial  method  of  briefly  writing  out  our  mental 
operations  when  we  solve  mathematical  problems,  and  as  such, 
it  may  be  extended  and  applied  to  almost  every  branch  of  the 
mathematics ;  and,  therefore,  the  value  of  this  science  cannot 
be  over  estimated. 

The  three  following  problems  are  extremely  simple  when 
algebraic  language  is  applied,  but  would  be  rather  difficult  by 
common  language. 

5.  On  a  certain  day,  a  merchant  jmid  out  ^2500  to  three  men, 
A,  B,  and  C;  he  paid  to  A  a  certain  sum,  to  B  $500  less  than 
the  sum  paid  A,  and  to  C  he  paid  $900  moi'e  than  to  A.  He- 
quired  the  sum  paid  to  each. 

Let       ....  x=  the  sura  paid  to  A. 

Then    ....     a:— 500=    "      *'  '«      B. 

And     ....     a;-|-900=    •*      *'  "      C. 


By  addition,  .     .  32-+400=2500,  the  whole  sum  paid. 


14  ELEMENTARY  ALGEBRA. 

Subtracting  400  from  both  members— that  is,  equals  from 
equals — and  we  have         3a;=2100. 

Take  J  of  each  member,  or  divide  both  sides  by  3  and 
ar=700,  the  sum  paid  to  A. 

From  this  we  determine  that  $200  was  paid  to  B,  and 
$1600  to  C. 

In  this  example  the  sign  before  500  shows  what  is  to  be 
done  with  that  quantity ;  it  shows  that  it  is  to  be  taken  out, 
and  accordingly  it  was  taken  out,  diminishing  the  900  to  400. 

This  operation  is  called  addition,  but  it  is  algebraic  addition, 
that  is,  writing  the  quantities  according  to  their  signs,  and 
finding  the  result. 

6.  It  is  required  to  divide  the  number  99  into  Jive  suck  parts 
thai  the  first  may  exceed  the  second  by  3,  be  less  than  the  third  by 
10,  greater  than  the  fourth  by  9,  and  less  than  the  fifth  by  16. 


Let    ...     . 

x=  the  first  part; 

then  .     .     .     . 

.    X —  3=  the  second. 

x-\-\0=  the  third. 

X —  9=  the  fourth. 

- 

ir+16=  the  fifth. 

Sum  is    .     .     . 

5a;+14,  but  the  sum  of 

99;  therefore, 

5ar+14=99. 

By  subtracting  14  from  both  members,  and  dividing  the 
remainders  by  5,  we  have  x=\l,  the  first  part;  and,  there- 
fore the  parts  are  17,  14,  27,  8,  and  33. 

7.  Divide  $1000  among  four  men,  giving  the  second  tidce 
as  much  as  the  first,  minus  $200;  the  third  double  the  sum  of 
the  second,  plus  $400;  the  fourth  three  times  as  much  as  ike 
first,  plus  $100.     Required  the  share  of  each. 

]jet     ....  rr=  the  sum  paid  to  the  1st, 

then   ....     2ar— 200=        **         "         "      2d, 
4.2r— 400+400=        "         "         *'      3d, 
3a;+100=        "         "         **      4th, 

The  sum  is  10a:— 100=  1 000. 


INTRODUCTION.  15 

By  adding  100  to  both  members  we  have 
10a:=1100 

By  adding  100  to  — 100  in  the  first  member  of  the  equa- 
tion, makes  0,  and  then  10a;  only  is  left  in  that  member,  which 
must  be  equal  to  1100,  or  a;=110,  the  sum  paid  to  the  first, 
and  the  several  sums  are  8110,  820,  8440,  and  8430. 

The  preceding  remarks  and  problems  serve  to  show,  only 
in  same  small  degree^  the  advantage  of  Algebra  over  common 
language,  and  the  learner  should  examine  every  problem,  and 
the  reason  of  every  step  in  the  process  of  its  solution,  until 
all  is  thoroughly  understood ;  then  he  will  have  no  difficulty 
in  solving  the  examples  that  follow  in  this  introduction.  But 
before  we  give  additional  problems,  let  us  call  the  student's 
mind  to  the  precise  idea  of  an 

EQUATION. 

An  equation  is  simply  what  the  word  implies ;  equality  as 
to  value,  weight,  or  measure  ;  and  can  be  best  understood  by 
comparing  it  to  a  pair  of  scales  delicately  balanced. 

The  halance  can  he  preserved  ly  adding  equal  weights  to 
hoth  sides;  by  taking  equal  weights  from  both  sides;  by  multiply- 
ing both  sides  by  the  same  number,  or  by  dividing  both  sides  by 
the  same  number,  or  by  talcing  like  roots  or  like  poivers  of  the 
weights  in  both  sides. 

The  object  of  working  an  equation  is  to  bring  the  unloiown 
quantity  to  stand  alone  as  one  member  of  the  equation,  equal 
to  known  quantities  in  the  other  member.  The  unknown 
quantity  thus  becomes  known;  and  we  may  do  anything  to 
accomplish  this  end,  that  the  nature  of  the  case  may  seem  to 
require,  only  taking  scrupulous  care  to  preserve  equality 
through  every  change. 

It  is  usual  to  represent  known  quantities  by  their  numerical 
values,  or  by  the  first  letters  of  the  alphabet,  as  a,  b,  c,  d,  <fec.; 
and  unknown  quantities  by  the  last  letters,  as  u,  i,  x,  y,  (fee. 


16  ELEMENTARY   ALGEBRA. 

AXIOMS. 

Axioms  are  self-evident  truths,  and  of  course  are  above 
demonstration ;  no  explanation  can  render  them  more  clear. 
The  following  are  those  applicable  to  Algebra,  and  are  the 
principles  on  which  the  truth  of  all  algebraical  operations 
finally  rests. 

Axiom  1.  If  the  same  quantity  or  equal  quantities  be 
added  to  equal  quantities,  their  sums  will  be  equal. 

2.  If  the  same  quantity  or  equal  quantities  be  subtracted 
from  equal  quantities,  the  remainders  will  be  equal. 

3.  If  equal  quantities  be  multiplied  into  the  same,  or  equal 
quantities,  the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same,  or  by  equal 
quantities,  the  quotients  will  be  equal. 

5.  If  the  same  quantity  be  both  added  to  and  subtracted 
from  another,  the  value  of  the  latter  will  not  be  altered. 

6.  If  a  quantity  be  both  multiplied  and  divided  by  another, 
the  value  of  the  former  will  not  be  altered. 

7.  Quantities  which  are  respectively  equal  to  any  other 
quantity  are  equal  to  each  other. 

8.  Like  roots  of  equal  quantities  arc  equal. 

9.  Like  powers  of  the  same  or  equal  quantities  are  equal. 
Now  suppose  we  have  the  following  equation 

xJ^a=^h 
in  which  x  is  the  unknown  quantity,  and  a  and  b  known  quan- 
tities. Before  x  can  become  known,  a  must  be  disensfacred 
from  it,  that  is,  a  must  be  subtracted  from  both  members. 
It  must  be  subtracted  from  the  first  member,  because  it  is  our 
object  to  have  x  stand  alone,  and  we  must  subtract  it  from 
the  other  member,  to  preserve  equality.  The  equation  then 
stands  x=b — a 

Here  we  find  the  quantity  a,  whatever  it  may  be,  cm  the  other 
side  of  the  equation,  with  the  contrary  sign. 


INTRODUCTION.  17 

Now  let  us  suppose  we  have  an  equation  like 
X — a-=b 

In  this  equation  we  perceive  that  x  is  diminished  by  a; 
therefore  to  have  the  single  value  of  x  we  must  add  a  to  the 
first  member ;  and,  of  course,  to  preserve  equality,  we  must 
add  a  to  the  second  member,  then  we  shall  have 
X — a+a=6-{-a 

But  — a  and  -]ra  destroy  each  other,  and  the  equation  is  in 
brief  x^=h-\-a 

Here,  also,  we  find  a  on  the  opposite  side  of  the  equation, 
with  its  sign  changed ;  and  from  these  investigations  we  draw 
the  following  rule  of  operation. 

Rule  . —  We  may  change  any  quantity  from  one  member  of 
an  equation  to  the  other,  if  we  change  its  sign. 

The  operation  itself  is  called  transposition. 

For  examples,  transpose  the  terms  so  that  the  unknown 
quantity  x  shall  stand  alone  in  the  first  member  of  the  follow- 
ing equations : 

x-\-c  —  d=4g     ....      Ans.  x==4g-\-d — c. 

x-\-3 — a-\-m=30     ....      Ans.  ar=30 — m-\-a — 3. 

SIMPLE  PROBLEMS  FOR  EXERCISES. 

1.  A  man  bought  a  saddle  and  bridle  for  45  dollars;  the 
saddle  cost  four  times  as  much  as  the  bridle.  What  was  the 
cost  of  each  ?  Ans.  Bridle  $9  ;  saddle  $36. 

2.  Three  boys  had  66  cents  among  them ;  the  second 
had  twice  as  many  as  the  first,  and  the  third  three  times  as 
many  as  the  first.     How  many  had  each  ? 

Ans.  1st  boy  had  11  ;  2d,  22 ;  3d,  33  cents. 

3.  Two  men  had  1 00  dollars  between  them,  and  one  had 
3  times  as  many  as  the  other.     How  many  had  each  ? 

Ans.  One  had  ^25,  th«  other  $75. 
K.  B.  This  last  problem  may  be  enunciated  thus  : 
Two  men  had  100  dollars  between  them ;  the  first  had 
one -third  as  many  as  the  other.     How  many  had  each  ?  . 
2 


18  ELEMENTARY   ALGEBRA 

4.  Three  men  had  880  dollars  among  them  ;  the  first  had 
~,  the  second  had  i  as  many  as  the  third.  Plow  many  had 
each  ?     Let  6x=  what  the  third  had. 

Ans.   1st  had  160  ;  2d,  240,  and  the  3d,  480  dollars. 

5.  There  are  three  numbers  which  together  make  72,  the 
second  is  twice  as  much  as  the  first,  and  the  third  is  as  much 
as  both  the  others.     What  are  the  numbers  ? 

A71S.  1st  is  12;  2d,  24;  3d,  36. 

6.  Two  men  built  90  rods  of  fence  in  3  days.  The  second 
built  twice  as  many  rods  in  a  day  as  the  first.  How  many 
rods  did  each  build  per  day  ?     Ans.  1st  built  10  rods,  2d,  20. 

7.  A  man  bought  3  oxen,  4  cows,  and  6  calves,  for  260  dol- 
lars. He  paid  twice  as  much  for  an  ox  as  he  did  for  a  cow, 
and  twice  as  much  for  a  cow  as  for  a  calf.  How  much  did 
he  give  for  each  ? 

Ans.  For  a  calf,  $10;  cow,  $20  ;  and  for  an  ox,  840. 

8.  A  man  bought  a  boat  load  of  flour  for  132  dollars,  one- 
half  at  5  dollars  per  barrel,  the  other  half  at  6  dollars  per 
barrel.     How  many  barrels  did  the  boat  contain  ? 

Let  x=  half  the  number  of  barrels.  Ans.  24. 

9.  A  boy  bought  an  equal  number  of  apples,  oranges  and 
pears,  for  96  cents :  the  apples  at  3  cents  apiece,  the  oranges 
at  4,  and  the  pears  at  5.  How  many  of  each  kind  did  he 
buy?  Ans.  8. 

10.  Two  men  bought  a  carriage  for  86  dollars;  one  paid 
five  times  as  much  as  the  other,  and  26  dollars  more.  What 
did  each  pay?  Ans.  One  paid  10,  the  other  76  dollars. 

11.  If  from  6  times  a  certain  number  we  subtract  24,  the 
remainder  will  be  196.     What  is  the  number  ?  Ans.  44. 

12.  To  the  double  of  a  certain  number,  if  we  add  18,  the 
sum  will  be  96.     What  is  the  number?  Ans.  39. 

1 3.  What  number  is  that  whose  double  exceeds  its 
78  ?     Let  2.v=  the  number. 

1 4.  A  man  had  six  sons,  to  whom  ho  gave  T 20  dollars,^givin^ 


INTRODUCTION.  19 

to  eacli  one  4  dollars  more  than  to  his  next  younger  brother. 
How  many  dollars  did  he  give  to  the  youngest  ?      Ans.  f  10. 

15.  Three  men  received  65  dollars  ;  the  second  received  5 
dollars  more  than  the  first,  and  the  third  10  dollars  more 
than  the  second.     What  sum  did  the  first  receive  ? 

Ans.  $15. 

16.  A  man  paid  a  debt  of  29  dollars,  in  three  different 
payments ;  the  second  time  he  paid  3  dollars  more  than  at 
first,  and  the  third  time  he  paid  twice  as  much  as  at  the  sec- 
ond time.     What  was  the  amount  of  his  first  payment  ? 

Ans.  $5. 
w.  A  man  bought  6  pounds  of  coffee,  and  10  pounds  of 
tea,  for  360  cents,  giving  20  cents  a  pound  more  for  the  tea 
than  for  the  coffee.     What  was  the  price  of  the  coffee  ? 

Ans.  10  cents. 

1 8.  A  man  bought  6  barrels  of  flour,  and  4  firkins  of  butter, 
for  68  dollars.  He  gave  2  dollars  more  for  a  barrel  of  flour 
than  for  a  firkin  of  butter.     What  was  the  price  of  flour  ? 

Ans.  $7.60. 

19.  A  pound  of  coffee  cost  5  cents  more  than  a  pound  of 
sugar,  and  for  3  pounds  of  sugar  or  for  2  pounds  of  coffee 
you  must  pay  the  same  sum.     What  is  the  price  of  sugar  ? 

Ans.   10  cents. 

20.  A  person  in  market  selling  apples,  peaches,  and 
oranges,  asked  1  cent  more  for  a  peach  than  for  an  apple, 
and  2  cents  more  for  an  orange  than  for  a  peach,  and  the 
prices  were  such  that  10  apples  and  5  peaches  cost  as  much 
as  5  oranges.     What  was  the  cost  of  an  apple  ?     Ans.   1  cent. 

21.  One-half  of  a  post  stands  in  the  mud,  one-third  in  the 
water,  and  the  remainder,  which  is  3  feet,  is  above  the  water. 
What  is  the  whole  length  of  the  post  ?  Ans.   18  feet. 

2S.'  One-third  and  one-half  of  a  sum  of  money,  and  ten 
dollars  more,  make  the  whole  sum.     What  is  the  sum  ? 

Ans.  60  dollars. 


20  ELEMENTARY  ALGEBRA. 

23.  Divide  25  cents  between  two  boys,  and  give  one  four 
times  as  mucli  as  the  other.     Required  the  share  of  each. 

Ans.  5  and  20  cents. 

24.  Divide  15  cents  between  two  boys,  and  give  one  double 
of  the  other.     Required  the  share  of  each. 

Ans.  5  and  10  cents. 

Similar  problems  to  the  preceding  might  be  framed  indefi- 
nitely, but  it  would  be  improper  to  propose  any  that  involve 
any  difficulty  until  the  pupil  is  better  prepared  to  meet  diffi- 
culties. We  only  give  the  preceding  to  convince  the  learner 
that  he  can  find  real  utihty  in  the  science ;  but  before  he  can 
go  into  the  subject  to  advantage,  he  must  learn  the  nature  of 
algebraic  expressions,  and  acquire  the  art  of  adding,  subtract' 
ing,  multiplying,  and  dividing  algebraic  quantities,  both  whole 
and  fractional. 

We  now  assure  the  young  beginner  that  we  will  conduct 
him  through  the  elements  of  this  science  with  as  little  delay 
and  trouble  as  possible ;  and  neither  remarks  nor  examples 
will  be  given  which'are  not,  in  the  judgment  of  the  author, 
essential  to  the  progress  of  the  pupil. 

With  this  assurance  we  close  this  introduction,  and  com- 
mence Algebra,  by  giving  more  extended  definitions  of 
terms. 


% 


ALGEBRA 


DEFINITION   OF   TERMS. 

The  signs  for  addition,  subtraction,  multiplication,  division, 
and  equality,  liave  already  been  explained.  We  have  also 
explained  coefficient  and  vinculum.  The  word  coefficient  can 
hardly  be  understood  by  a  mere  definition.  It  means  any 
factor  connected  with  another,  and  may  be  simple  or  com- 
pound, thus,  ax;  a  is  the  coefficient  of  x,  and  in  the  term 
3ax,  3a  is  the  coefficient  of  x,  and  3  is  the  coefficient  of  ax. 

1.  When  a  letter  stands  alone,  as  b,  y,  or  any  other  letter, 
one  or  unity  may  be  considered  its  coefficient.  In  the  expres- 
sion (3a4-25 — c)x,  (3a-|-25 — c)  is  a  compound  coefficient  to 
X.  It  is  also  a  factor,  and  x  is  another  factor.  The  word 
factor  has  the  same  signification  as  in  Arithmetic. 

2.  When  we  wish  to  note  that  two  quantities  are  unequal y 
we  write  this  sign  ^  between  them.  The  opening  of  the 
sign  is  always  put  toward  the  greater  quantity,  thus,  a'^hy 
signifies  that  a  is  greater  than  6,  and  a<^h  shows  that  h  is 
greater  than  a. 

3.  When  we  indicate  the  multiplication  of  numbers,  with- 
out actually  performing  the  multiplication,  we  must  write  the 
sign  X  or  •  between  them,  as  5X  5,  or  3'4,  because  the  mul- 
tiplication could  not  be  understood  without  the  sign ;  but 
when  we  have  letters  in  place  of  numbers,  we  may  omit  the 
sign  and  write  ah,  in  place  of  «  X  6 ;  ahx  in  place  of  a*h*x,  <fec 


22  ELEMENTARY  ALGEBRA. 

4.  When  factors  are  equal,  and  each  equal  to  a,  the  pro- 
duct of  four  such  factors  is  aaaa;  but  in  place  of  this,  we 
may  write  a^  which  signifies  that  a  is  taken  four  times  as  a 
factor ;  a?  indicates  a  product  which  is  composed  of  a  taken 
seven  times  as  a  factor ;  a;°  indicates  a  product  in  which  x  is 
taken  as  many  times  as  a  factor  as  there  are  units  in  n. 

5.  The  smcdl  number  thus  written  to  the  right  of  a  letter, 
(or  to  any  quantity)  and  a  little  above,  is  called  its  exponent. 
Expmierds  may  be  either  whole  numbers  or  fractions;  but 
this  will  be  explained  hereafter. 

When  an  exponent  is  a  whole  number,  it  indicates  the 
power  of  the  factors,  or  quantity  to  which  it  is  attached. 
When  it  is  a  fraction,  it  indicates  a  root  of  the  quantity, 
thus,  c^  indicates  the  square  or  second  root  of  a  ;  a^  indicates 
the  third  root  of  a,  and  a*  indicates  the  fourth  root  of  a,  &c. 
Formerly,  the  sign  of  a  root  was  indicated  by  the  radical 
sign  J;  and,  for  some  purposes,  this  sign  is  still  in  use. 
Whenever  it  is  used,  it  is  placed  to  the  left  of  the  quantity ; 
thus,  Ja.  The  nwmher  of  the  root  is  denoted  by  a  little 
figure  placed  over  the  radical  sign ;  unless  it  is  the  second 
root,  when  the  figure  2  is  omitted.     Thus, 

J  a  is  the  second  or  square  root  of  a. 

II a  is  the  third  or  cube  root  of  a. 

lla\s  the  nt\\  root  of  a. 

Ja-\-x  is  the  second  root  of  (a-j-a;). 
It  must  be  remembered,  that  if  the  radical  sign  is  to  affect 
more  than  one  factor,  the  vinculum  must  be  used  with  it; 
thus,  Jba,  Jab.  ■^. 

6.  The  number  of  literal  factors  which  enter  into  any  term, 
is  the  degree  of  that  term ;  ah  is  of  the  second  degree,  cth  of 
the  third,  a6V  of  the  fifth.  In  general,  the  degree  of  an 
algebraic  term  is  found  by  taking  the  sum  of  the  exponents 
of  the  letters  which  enter  into  that  term.  An  algebraic  quan- 
tity which  has  all  its  terms  of  the  same  degree,  is  said  to  be 


DEFINITION  OF  TERMS.  23 

homogeneous:  4a'^+2a^5^ — '^ah(?-\-'b^c,  is  therefore  homogene- 
ous, and  of  the  fifth  degree.  * 
Similar  terms,  are  those  which  contain  tlie  same  letters  in 
the  same  powers,  2a^b  and  5a^b  are  similar.  But,  3ab^  and 
Sab  are  not  similar  terms,  for  the  letters,  although  the  same, 
are  not  in  the  same  power. 

7.  Simple  Quantities,  are  those  which  consist  of  one  term 
only.     As  3a,  or  5ab,  or  6ab(^. 

8.  Compound  Quantities,  are  those  which  consist  of  two  or 
more  terms.     As  a-\-b,  or  2a — 3c,  or  a-{-2b — 3c. 

9.  And  when  the  compound  quantity  consists  of  two  terms 
it  is  called  a  Binomial,  as  a-^b  ;  when  of  three  terms,  it  is  a 
Trinomial,  as  a-\-2b — 3c ;  when  of  four  terms,  a  Quadrino- 
mial,  as  2a — 3b-\-c — 4c? ;  and  so  on.  Also,  a  Multinomial  or 
Polynomial,  consists  of  many  terms. 

10.  A  Residual  Quantity,  is  a  binomial  having  one  of  the 
terms  negative.     As  a — 2b. 

11.  Positive  or  Affirmative  Quantities,  are  those  which  are 
to  be  added,  or  have  the  sig-n  +•  As  a  or  -\-a,  or  ab:  for 
when  a  quantity  is  found  without  a  sign,  it  is  understood  to 
be  positive,  or  as  having  the  sign  4"  prefixed. 

12.  Negative  Quantities,  are  those  which  are  to  be  sub- 
tracted.    As  — a,  — 2ab,  or  — 8c^ 

13.  Like  Signs,  are  either  all  positive  (+),  or  all  negative 

14.  Unlike  Signs,  are  when  some  are  positive  (+),  and 
others  negative  ( — ). 

15.  The  reciprocal  of  any  quantity  is  unity  divided  by  that 

quantity.     Thus,  i  is  the  reciprocal  of  4,  -  is  the  reciprocal 

a 
of  a,  and  so  on,  of  any  other  quantity. 

16.  The  same  letter,  accented,  is  often  used  to  denote 
quantities  which  occupy  similar  positions  in  different  equa- 
tions or  investigations.     Thus,  a,  a',  a",  a'",  represent  four 


24  ELEMENTARY  ALGEBRA. 

different  quantities ;  of  wliicli  a!  is  read  a  prime  ;  a"  is  read 
a  second  ;  a!"  is  read  a  third,  and  so  on. 

That  the  pupil  may  imbibe  or  catch  the  true  spirit  of  an 
algebraic  expression,  we  give  the  following  exercises  in  con- 
verting common  arithmetical  operations  into  algebraic  ex- 
pressions, and  finding  the  value  of  each  under  different 
suppositions. 

Express  in  algebraic  langiuige  the  prodtict  of  three  times  a 

into  X,  diminished  by  c,  and  the  remainder  divided  by  b. 

.        Zax — c 

Ans.  . 

b 

What  is  the  value  of  this  expression  when  a =2,  x=S, 
«=4,  and  5=2  ?  Ans.  7. 

What  is  its  value  when  a=3,  x=5,  c=9,  and  b=3  ? 

Ans.  12. 
Express  in  algebraic  language  3  tim£S  the  square  of  a,  dimin- 
ished by  2b,  and  the  difference  divided  by  c.         An^.  . 

c 

What  is  the  numerical  value  of  this  expression,  when  a=5, 
5==10,  and  c=5?  Ans.  11. 

Wllatwhena=10,  5=7,  and  c=20?  .  .  Ans.  \^j\. 
"What  when  a=  9,  5=0,  and  c=   1  ?       .     .     Ans.     243. 

What  when  a=   1,  5=1,  and  c=  |?       .     .     Ans.         2. 

Write  the  followmg:  Qa  diminished  by  x,  the  diflPerence 
increased  by  the  square  root  of  c,  and  the  whole  multiplied 
by  5.  Ans.  (6a — x-]rc^)b. 

What  is  the  value  of  this  expression,  when  a=3,  a;=18, 
c=  16,  and  5=2?  Ans.       8. 

What  when  «=6,  rr=9,  c=9,  and  5=7  ?  Ans.  210. 

What  is  the  value  of  the  expression  a^-\-2>ah — c^,  when 
a=:6,  5=5,  and  c=4  ?  Ans.  1 10. 

With  the  same  value  to  a,  b,  and  c,  what  is  the  value  of 
the  expression,  2a^ — 3a'^5-|-c^  ?  Ans,  — 44. 

What  of  the  expression  a^(a-\-b) — 2a5c?  Ans.     166. 


ADDITION.  ^ 

a? 
What  is  the  value  of    -r—--\-c^'l ^«s.  28. 


What  is  the  value  of  a^-^Jb^—<ic'>    ....     AJIk2,b. 


w 


SECTION    I. 


ADDITION. 

(Art.  1.)  Addition  in  Algebra  is  connecting  quantities  to- 
gether by  their  proper  signs. 

Here  the  pupil  should  call  to  mind  the  fact  that  unlike 
quantities  cannot  be  added  together.  For  instance,  it  would 
be  an  absurdity  to  add  dollars  to  yards  of  cloth,  and  so  of 
any  other  unlike  quantities ;  but  dollars  can  be  added  to  dol- 
lars, yards  to  yards,  &c.;  so  in  Algebra,  a  may  be  added  to 
a,  making  2a,  or  any  number  of  a's  may  be  added  to  any 
other  number  of  a's  by  uniting  their  coefficients  ;  but  a  can- 
not be  added  to  b  or  to  any  other  dissimilar  quantity :  we 
can  write  a-\-b,  indicating  the  addition  by  the  sign  making  a 
compound  quaiitity. 

(Art.  2.) Addition  in  Algebra  may  be  divided  into  three 
cases :  the  first,  when  the  quantities  are  alike  and  their  signs 
alike ;  a  second,  when  the  quantities  are  alike  and  the  signs 
unlike  ;  and  the  third,  when  the  quantities  are  unlike. 

To  discover  a  rule  for^ase  1st,  we  propose  the  following 
problem : 

On  Monday  a  merchant  sent  to  a  steamboat  17  barrels  of 
flour  and  9  barrels  of  pork ;  on  Tuesday  he  sent  7  barrels 
of  flour  and  10  of  pork;  on  Wednesday,  20  barrels  of  flour 


26  ELEMENTARY  ALGEBRA. 

and  6  of  pork ;  on  Thursday,   10  barrels  of  flour  and  10 
of  pork.     How  many  barrels  of  each  has  he  sent  ? 

Way  write  it  thus  : 
17  barrels  of  flour  +    9  barrels  of  pork 

'J         it  <(        it       JL.    1Q         ti  tt       it 


20 

tt 

it 

tt 

+    6 

it 

it 

it 

10 

it 

it 

tt 

+  10 

it 

it 

it 

64  barrels  of  flour  +35  barrels  of  pork 

Now  let  b  represent  a  barrel  of  flour,  and  p  represent  a 

barrel  of  pork,  then  in  place  of  writing  out  the  words,  we 

write  175+  9p 

75+1  Op 

205+  6p 

105+10;; 

Sum  is 545+35/; 

From  this  example  we  perceive  that  to  add  together  simi- 
lar quantities,  we  have  only  to  add  their  numeral  coefiicients, 
like  simple  numbers  in  Arithmetic. 

Hence,  the  following  rule  will  meet 

Case  1 .  When  the  quantities  are  similar  and  the  siffns 
alike,  add  the  coefficients  together,  and  set  down  the  sum  ;  after 
which  set  t/ie  common  letter  or  letters  of  the  like  quantities,  and 
prefix  the  common  sign  +  or  — . 


EXAMPLES. 

(0 

(2) 

(3) 

(4) 

,        (5) 

3a 

—  3bx 

bmj 

3a+  25—  5c 

4a5—  2crf 

9a 

—  bbx 

2bxy 

6a4-  65—     c 

7a5—    cd 

5a 

—  Abx 

bbxy 

7a-tll5—  8c 

\5al>—  2cd 

12a 

—  25a; 

—  75a; 

bxy 
Zbxy 

a+     5—  3c 

ab—12cd 

a 

16a+205— 17c 

27  ab— lied 

2a 

—     bx 

Qbxy 

32a        —225*        1850^ 


ADDITION.  27 

(6)  (7)  (8) 

3a-t-  3ax-\-  c  4a6+3^ — 2b  lOy —     x-\-     k 

la-\-  5ax-\-5c  lah-{-  x — 36  ly —     x^  2>h 

10a-|-  lax-\-2c  121aZ>-f  2a;—  h  2>y~  2x-\-  Ih 

2a+10aa;+4c  99a64-  a; —  b  y—\Ox-\-\Qh 

Sum 

(Art.  3.)  Like  quantities,  of  whatever  kind,  whether  of 
powers  or  roots,  may  be  added  together  the  same  as  more 
simple  quantities. 

Thus  3a2  and  Sa^  are  1  la^  and  lW-^W=\Qb^.  No  matter 
what  the  quantities  may  be,  if  they  are  only  alike  in  kind.  Let 
the  reader  observe  that  2(aH-5)+3(a-|-6)  must  be  together 
5(a4-6),  that  is,  2  times  any  quantity  whatever  added  to  3 
times  the  same  quantity,  must  be  five  times  that  quantity. 
Therefore,  ^Jx-\-y-{-3jx-\-y=7jx-\-y,  for  ^x-{-y,  which 
represents  the  square  root  of  ic+y,  may  be  considered  a  single 
quantity. 

To  illustrate  these  remarks  we  give  the  following 


EXAMPLES. 

4(a—x) 

{^-\-y)         Ja-\-^ 

6{a^-c) 

7(a^x) 

3(a;+2/)            Qja^x 

{a^-c) 

I0(a—x) 

20{x-\-y)          \2,Ja+x 

l{a^-c) 

Sum      2\{a—x)         24(a;+y)         ^O^a-^x 

(Art.  4.) — Case  2.   When  the  quantities  are  similar  and 
tJie  siyns  unlike,  we  have  the  following  rule  for  addition.* 


*  In  this  rule,  the  word  addition  is  not  very  properly  used ;  being 
much  too  limited  to  express  the  operation  here  performed.  The  busi- 
ness of  this  operation  is  to  incorporate  into  one  mass,  or  algebraic 
expression,  different  algeltraio  quantities,  as  far  as  an  actual  incorpora- 


28  ELEMENTARY  ALGEBRA. 

Rule. — Add  the  affirmative  coefficients  into  one  sum  and 
the  negative  ones  into  another,  and  take  their  difference  with  the 
sign  of  the  greater,  to  which  affix  the  common  literal  quantity. 


Sum 


EXAMPLES    FOR 

PRACTICE 

(1) 

(2) 

(3) 

^5a 

+3tt;r« 

+  8af^-f  3y 

+4a 

+4aa:2 

— '5^+4^ 

+6a  , 

—8(^2 

—16x'-\-5g 

~3a 

—Qax^ 

+  3a:«— 7y 

+  a 

-\-Sa3^ 

+  2:ir'— 22/ 

+3a 

—^aj? 

—  82^=^4- 3y 

(4) 

(5) 

(6) 

—  So" 

+  3^y 

+4ai+  4 

—  bd' 

4-  9iy 

— 4a5-}-12 

—  lOa^ 

— lo^y 

-|-7a5_14 

4-10^2 

—\Wf 

+  a5  +  3 

+  14a2 

—  2uy 

—Bab  —10 

7.  Add  2xy—2a\  JSa^-f-ary.  a^-fay,  4a2_3;ry,  2xg—2a\ 

Ans.  4a^-{-3xy. 

8.  Add  8aV — 3xy,      Box — 5xy,      9xg — box,      2a^a^-\-xj/, 
Sax — 3xy.  Ans.   \0c^.t^-\-5ax — xy. 

9.  Add   3^2—1,    6am—2m:^-\-4,  •7—Sam-{-2m\  and   6m^ 
+2am-\-l.  Am.  Om'+ll. 


tion  or  union  is  possible ;  and  to  retain  the  algebraic  marks  for  doing 
it,  in  cases  where  the  former  is  not  possible. 

By  using  the  word  united  in  place  of  the  word  added,  the  reason  of 
the  rule  will  become  obvious. 

Thus  3a  united  to  —  a  makes  2a 

Ix  united  to  —2a;  makes  hx 


ADDITION.  29 

10.  Add  12a—lSah-{-16ax,  8— 4m+2y,  —Sa+TaS^^- 
12y — 24,  and  lab — 16ax-i-4m. 

Ans.  Qa—Qah-\-\Ay-\-lalr—\Q. 

1 1 .  Uriite  4a^b — Qarb — 9a^b-\- 1 1  a/^b  into  one  term  if  possible. 

Ans.  — 2a^b. 

12.  Ufiite  1ab(? — ab<? — lab^ — 8a5c^-|-9«5c^  into  one  term. 

Ans.  0. 

13.  Add  3a(a4-6),  la{a-\-b),  —ba{a-\-b),  and  ^a{a-^b), 

Ans.  8a(a+6). 

14.  Add  7(6.r4-y— 2)^  — 8(6a:+y— 2)2,  (i^x-^y—zf,  and 
3(6ar+y— 2)'.  ^tzs.  3(6x4- y— 2)2. 

15.  Add  3a5+4ii(62/+^),  — 8a5— 9a(62/-}-Z>),  12a64-13a 
(63/+5),  «i+a(6?/+5),  and  7a^4- 6a(6y4-5). 

^ws.  15a64-15a(%+^)- 
C  A  s  E  3 . —  When  the  quantities  are  unlike  and  the  signs 
ALIKE  or  UNLIKE,  we  havc  the  following  rule  to  unite,  or  rather ' ' 
to  reduce  and  condense  the  quantities.  , 

R  u  L  E. —  Collect  together  all  those  terms  that  are  similar,  by 
uniting  their  coefficients,  as  in  the  former  cases  :  then  lorite  the  dif- 
ferent sums,  one  after  another,  udth  their  proper  signs. 

N.  B.  It  is  immaterial  what  quantity,  in  an  aggregate  sum, 
stands  first,  for  the  whole  of  a  thing  is  equal  to  the  sum  of 
all  its  parts,  whatever  part  may  be  first  written.  Thus, 
ax-\-l}y-{-c  is  the  same  sum,  whichever  terai  stands  first. 

EXAMPLES. 

1.  Add.  3a f,  — 2xy^y    — 3y^x,  — Qct^y,  and  2xy\ 

r3af — 2xy^ 
These  terms  may  be  arranged  thus :  <        — 3xy^ — 8s^y 

i  +2a-f 
Sum 3ay — 3xy — Sx'^y 

2.  Add  together  15a^—Qbh-{-32ah'^—12bc,  19Pc—4a^-\- 
lla^c^-\-2bc,  a^— 29aV— 12^»''^c+56c,  and  9a^c'^—14bc-\-b''c. 


30  ELEMENTARY  ALGEBRA. 

a^— 12Zi2c— 29aV-i-  bbc 
+     bh-[-   9aV— 145c 


12a2    *  *  4-23aV— 19k,  the  ans. 

3.  What  is  the  sum  of  6a5-{-125c— 8cc?,t3cc?— 7a5— 95c, 
and  12cd—2a7}—5bc^  Ans.  7cd—2al)—2bc. 

4.  What  is  the  sum  of  5j^—7jbc-\-M,  ^Jd>^^jTc 
-12c?  and  1  Jab-\-2>Jbc-YM'>        Ans.  15j'^-\-4jb^-\-5d. 

5.  Add  72ax^—8af,  —SQaa;'^—Say*+7af,  S-\-l2ay\ 
'-6af-\-12—34ax'^-i-5a7/—9a2/\  Ans.  — 2a/+20. 

Add  a-\-b  and  3a — 5b  together. 

Add  6x—5b-\-a-\-8  to  -_5a— 4a;+45--3. 

Add  a4-25— 3c— 10  to  35— 4a+5c-|-10  and  65— c. 

Add  3a-f5— 10  to  c—d—a  and  — 4c+2a— 3&--7. 

Add  Sa^-\-2b^—c  to  2a5— 3a2+Jc— 5.  C 

(Art.  5.)  Let  it  be  strictly  observ^ed,  that  when  we  add  sim- 
ilar quantities  together,  as  Sx,  4x,  and  10a;,  we  perform  it  by- 
writing  the  coefficients  in  one  sum,  17,  and  writing  the  x,  or 
the  quantity,  whatever  it  may  be,  afterward,  making  in  this 
example  17ar.  As  principles  never  change,  we  must  do  the 
same  thing  when  the  coefficients  are  literal;  thus,  the  sum  of 
ax,  bx,  and  ex  must  be  [a-\-b-\-c)x,  and  ax — x  may  be  written 
{a—\)x. 

EXAMPLES. 

(1)  (2) 

Add ax-\-by^  ay-{'cx 

2ca:+3ay*  Say-\-2cx 

Adx-\-li/  4y  -]rQx 


Sum    (a-i-2c+4c?)a.'-l-(5+3a+7)2/2     (4a-l-4)y-f  (3c-f  6).r 


ADDITION.  ai 

(3)  (4) 

AM     .     .     .     3a;H-2d-2/  (tx-{-7y 

hx  -{-cxy  lax — 3y 

(a-\-h)x-\-9.cdxy  — 9.x  +4y 

Sum  {a-{-9b-\-2>)x-\-{'^cd-\-c-^9)ry      (8a— 2)a;-|-8y 

5.  Add   8ai:+2(^+a)4-3i,  Qax-^Q{x-\'a)—%,  and   11a; 
4-6^— 7aa;— 8(a;-fa).  ^tw.   lOax-^Ux. 

6.  Add  {a-\-b)Jx  and  (c+2a — ^J^o:  together. 

^ns.  (c+3a)^ar. 

7.  CTTii^e  3ax-{-7ax — 4ax — lx-\-3hx-{-4x,  as  far  as  possible, 
and  find  the  sum  total  of  the  coefficient  of  x. 

Ans.  6ax-\'2bx-\-4x\ 
The  sum  total  of  the  coefficient  of  x  is  (6a-|-26-f-4),  and 
the   sum  total   of   the   whole    expression    may   be   written 
(6a+254-4>. 


SUBTRACTION. 

(Art.  6.)  Subtraction  in  Algebra  is  not,  in  all  cases,  taking 
one  quantity  from  another :  ii  is  finding  the  difference  between 
two  quantities. 

What  is  the  difference  between  12  and  20  degrees  of 
north  latitude  ?  This  is  subtraction.  But  when  we  demand 
the  difference  of  latitude  between  6  degrees  north  and  3 
degrees  south,  the  result  appears  like  addition,  for  the  differ- 
ence is  really  9  degrees,  the  sum  of  6  and  3.  This  example 
serves  to  explain  the  true  nature  of  the  sign  minus.  It  is 
merely  an  opposition  to  the  sign  plus  ;  it  is  counting  in  another 
direction^  and  if  we  call  the  degrees  north  of  the  equator  jy/w5. 


3a  ELEMENTARY  ALGEBRA. 

we  must  call  those  south  of  it  minus,  taking  the  equator  as 
the  zero  line. 

So  it  is  on  the  thermometer  scale,  the  divisions  above  zero 
are  called  pltcs,  those  below  minus.  Money  due  to  us  may  be 
called  plus,  money  that  we  owe  should  then  be  called  minus, 
— the  one  circumstance  is  directly  opposite  in  effect  to  the 
other.  Indeed,  we  can  conceive  of  no  quantity  less  than  nothing^ 
as  we  sometimes  express  ourselves.  It  is  quantity  in  oppo- 
site circumstances  or  counted  in  an  opposite  direction ;  hence 
the  difference  or  space  between  a  positive  and  a  negative  quantity 
is  their  apparent  sum. 

As  a  further  illustration  of  finding  diflferences,  let  us  take 
the  following  examples,  which  all  can  understand : 

From     .    16  16  16  16  16  16 

Take      .12  8  2  o  — 2  -—4 

Differ.    .      4  8  14  16  Ts  20 

Here  the  reader  should  strictly  observe  that  the  smaller 
the  number  we  take  away,  the  greater  the  remainder,  and 
when  the  subtrahend  becomes  minus,  its  numeral  value  must 
be  added. 

(Art.  7.)  We  cannot  take  a  greater  quantity  from  a  less; 
but  we  can,  in  all  cases,  Jind  the  difference  between  any  two 
quantities,  and  if  we  conceive  a  greater  quantity  taken  from 
a  less,  the  diflference  cannot  be  positive,  but  must  be  negative, 
i.  e.  minus. 


EXAMPLES. 

From 

12 

12               12 

12 

12 

12 

Take       . 

30 

20           16 

12 

10 

6 

Diflfer.    . 

—18 

—8         —4 

0 

2 

6 

(Art.  8.)  When  we  take  any  quantity  from  zero,  the  differ- 
ence wUl  he  the  same  quantity  with  its  sigri  changed,  as  will  be 
obvious  from  the  following  examples : 


SUBTRA.CTION. 

{ 

From     . 

.     10a 

3a             0 

0 

~5a 

Take      . 

.     11a 

6a             6a 

—6a 

—6a 

Differ.    . 

.    — a 

— a         — 6a 

■^6a 

a 

(Art.  9.)  Unlike  quantities  cannot  be  written  in  one  sum, 
(Art.  1,)  but  must  be  taken  one  after  another  with  their 
proper  signs :  therefore,  the  difference  of  unlike  quantities 
can  only  be  expressed  by  signs.  Thus,  the  difference  be- 
tween a  and  b  is  a — b,  a  positive  quantity  if  a  is  greater  than 
b,  otherwise  it  is  negative.  From  a  take  b — c,  (observe  that 
they  are  unlike  quantities.) 

OPERATION. 

From a-f-O+O 

Take 0-\-b—c 

Remainder,  or  difference,  a — b  -\rc 

This  formal  manner  of  operation  may  be  dispensed  with ; 
the  ciphers  need  not  be  written,  and  the  signs  of  the  subtra- 
hend need  only  be  changed. 

From  the  preceding  observation,  we  draw  the  following 

GENERAL  RULE   FOR   FINDING    THE    DIFFERENCE   BETWEEN   ALGE- 
BRAIC  QUANTITIES. 

Rule  . —  Write  the  terms  of  the  suUrahend,  one  after  another, 
with  their  signs  changed  ;  and  then  unite  terms,  as  far  as  j^ossl- 
ble,  by  the  rules  of  addition. 

Or  we  may  give  the  rule  in  the  following  words  : 
Conceive  the  signs  in  the  subtrahend  to  be  changed,  and  then 
proceed  as  in  addition. 

EXAMPLES. 

(1)  (2)  (3) 

From     .     .     .      4a-j-2.c — 3c  3aa;+2y  a-\-h 

Take      .     .     .        a-\-^x — 6c  xy — 2y  a — b 

Remainder,     .      3a,— 2a;4-3c     3aa: — xy-\-^y  26 


34  ELEMENTARY  ALGEBRA. 

(4)  (6)  (6) 

From     .     .        ^^—^x-\-f  7a-}-2—5c        ix-\-\y 

Take      .     .      — a? — Ax-\-a  — a*\-'2.-\-  c        '^x — \y 

Rem.     .     .  2>x'-\-x^f—a  8a  *  —^c  'y 

(7)  (8)  , 

From     .     .     .     ^2^ — 2>xy-\-2'f-\-  c  ax-\-bx-\-cx 

Take      .     .     .      x" — Qxy-\-'2>y" — 2c  x-\-ax-\-hx 

Difference,     .      lx^-^2,xy—  y^-\-^c  Jc—\)x 

9.  Find  the  difference  between  ^xy — 20  and  — xy-\-^^' 

Ans.^9xy — 32. 

10.  Find  tlie  difference  between  '7arx-\-a  and  3a^x — 2a. 

Ans.  4a^x-j-3a. 

11.  Find  the  difference  between  — Bx — 2y-\-3  and  10a; 
— 3y+4.  Ans.  — 18^+y— 1. 

12.  Find  the  difference  between  6y^ — 2y — 5  and  — 83/^ 
— 5y4-12.  Am.  14?/+3y— 17. 

13.  From  13aV/+lla— Sa^-f-G^*, 

Take  '7a—5a'^-\-6b—10a'b^       Remainder,  23a^b''-{-4a. 

14.  From  3a+b+c—d—10, 

Take  c-{-2a—d.  Rem.  a4-^— 10. 

15.  From  3a+^.-l-c— c?— 10, 

Take  5— 19+ 3a.  Rem.  c— c/+9. 

16.  From  2ab-\-b^—4c-\-bc—b, 

Take  3a:'—c-{-b\  Rem.  2ab—3c+bc— 30"— b. 

17.  From  a!'+3b^c+ah^—abc, 

Take  P-\-ab^—abc.  Rem.  a"+3b''c—b\ 

(Art.  10.)  From  a  take  5.  The  result  is  a — b.  The 
minus  sign  here  shows  that  the  operation  has  been  performed  : 
b  was  positive  before  the  subtraction ;  changing  the  sign  per- 
formed the  subtraction;  so  changing  the  sign  of  any  other 
quantity  would  subtract  it. 


SUBTRACTION  S!^ 

18.  From  3a  take  (ab-j-x — c — ?/),  considering  the  terms  in 
the  vinculum  as  one  term,  the  difference  must  be  3a — (ab-]rx 
— c — ^y),  but  if  we  subtract  this  quantity,  not  as  a  whole,  but 
term  by  term,  the  remainder  must  be  3a — ab — x-{-c-\-y. 

That  iSf  when  the  vinculum  is  taken  away,  all  the  signs  with- 
in the  vinculum  must  be  changed. 


EXAMPLES. 

1.  From  2>0xy,  take  (40a^— SS^+Sc— 4c?). 

Kern.  9,b^—\0xy—3c-\-Ad. 
From  3a2,  take  (3a — x-{-b).  Rem.  Sa^ — 3a-{-x—b. 

From  c? — a,  take  4a — y — 3a^- — 1. 

Rem.  a^ — a — (4a — y — 3a^ — 1). 

Or Aa^—5a-^y-\-\, 

From  a-\-b,  take  a — b. 
From  4a4-4&,  take  b-\-a. 
From  4a — 46,  take  3a-l-5J. 
From  8a — \9,x,  take  4a — 3x. 

(Art.  11.)  It  will  be  a  useful  exercise  for  the  mind  to  look 
at  the  principle  of  subtraction  in  Algebra,  through  the  medi- 
um of  equations. 

If  we  subtract  12  from  18,  the  remainder  will  be  6.  Here 
are  three  quantities. 

1 .  The  minuend       1 8 

2.  The  subtrahend   12 


3.  The  remainder 


In  all  cases,  the  remainder  and  the  subtrahend,  added  to- 
gether, must  equal  the  minuend.  N'ow  let  us  suppose  that 
we  do  not  know  the  value  of  the  remainder,  and,  therefore, 
represent  it  by  the  letter  R.  Then  by  the  nature  of  the 
case  we  have  72-|-12  =  18,  an  equation. 


36  ELEMENTARY  ALGEBRA. 

Taking  equals  from  equal  quantities,  that  is,  12  from  both 
members  of  the  equation,  we  have 
Ii=6 

Now  let  us  take  the  third  example  under  the  last  rule,  and 
call  its  remainder  H. 

Then  we  have       Ji-\-a — b=a-\-b 

Rejecting  a  from  both  members,  and  adding  h,  or  (what  is 
the  same  thing),  transposing  — 5,  (see  page  16),  and  we  find 
R=^2b 

Take  example  9,  and  we  have 

R—xy-{- 1 2 = Sary— 20 

By  transposition,  Ii=9xy — 32 

In  this  manner  we  ma?/  perform  all  the  examples  in  sub- 
traction ;  and  in  this  manner  perform  the  following  examples : 

From  2a-\-2b,  take  — a — b. 

From  ax-\-bx,  take  ax — bx. 

From     a-]rc-[-b,  take  a+c — b. 

From  Sx-\-2y-\-2,  take  5x-]-Sy-{-b. 

From  6a-\-2x-{-c,  take  5a-]rGx — 3c. 


MULTIPLICATION. 

(Art.  12.)  The  nature  of  multiplication  is  the  same  in 
Arithmetic  and  Algebra.  It  is  repeating  one  quantity  as 
many  times  as  there  are  units  in  another ;  the  two  quantities 
may  be  called  factors,  and  in  abstract  quantities  either  may 
be  called  the  multiplicand ;  the  other  of  course  will  be  the 
multiplier. 

Thus,  4X5.  It  is  indifferent  whether  we  consider  4  re- 
peated 5  times,  or  6  repeated  4  times  ;  that  is,  it  is  indififerent 


MULTIPLICATION.  S7. 

which  we  call  the  multiplier.  Let  a  represent  4,  and  b  repre- 
sent 5,  then  the  product  is  aXb;  or  with  letters  we  may  omit 
the  sign,  and  the  product  will  be  simply  ab. 

The  product  of  any  number  of  letters,  as  a,  b,  c,  d,  is  abed. 

The  product  of  x,  y,  z,  is  xi/z. 

In  the  product  it  is  no  matter  in  what  order  the  letters  are 
placed,  xy  and  yx  is  the  same  product. 

The  product  oi  axXby  is  axby  or  abxy.  Now  suppose  a=6 
and  5=8,  then  a5=48,  and  the  product  of  axXby  would  be 
the  same  as  the  product  of  6a;X8y  or  48a^.  From  this  we 
draw  the  following  rule  for  multiplying  simple  quantities, 
which  may  be  called 

Case  1 .  Multiply  the  coefficients  together y  and  annex  the  let- 
ters, one  after  another,  to  tlie  product. 

EXAMPLES. 

1.  Multiply  3a;  by  7a Product  21  oar. 

2.  Multiply  Ay  by  ^ab Product  l^aby. 

3.  Multiply  36  by  5c,  and  that  product  by  10a;. 

Ans.  \5Qbcx. 

4.  Multiply  Qax  by  \9,by  by  lad.         Ans.  504aaxydb. 

5.  Multiply  Sax  by  76  by  3y.  Ans.  63abxy. 

6.  Multiply  lOOaxy  by  lOa^cy  by  2.     Ans.  2000aabcxyy. 

In  the  preceding  examples  no  signs  were  expressed,  and  of 
course  plus  was  understood  as  belonging  to  every  factor; 
and  a  positive  quantity,  taken  any  number  of  times,  must  of 
course  be  positive. 

(Art.  13.)  As  algebraic  quantities  are  liable  to  be  affected 
by  negative  signs,  we  must  investigate  the  products  arising 
from  them.  Let  it  be  required  to  multiply  — 4  by  3,  that  is, 
repeat  the  negative  quantity  3  times,  the  whole  must  be  neg- 
ative, became  a  negative  quantity  taken  any  number  of  times 
must  be  negative.     Hence  minus  multiplied  by  plus  gives 


38  ELEMENTARY  ALGEBRA. 

minus,  — aXh  gives  — ah;  also  a  multiplied  by  — h  must 
give  — ah,  as  we  may  conceive  the  minus  h  repeated  a  times. 

!N"ow  let  us  require  the  product  of  — 4  into  — 3. 

In  all  cases  the  multiplier  shows  how  many  times  the  mul- 
tiplicand must  be  taken; — when  the  multiplier  is  plus,  it 
shows  that  the  multiplicand  must  be  added  to  zero  as  many 
times  as  there  are  units  in  the  multiplier; — when  the  mul- 
tiplier is  minus,  it  shows  that  the  multiplicana  muse  ue  suo- 
traded  from  zero  as  many  times  as  there  are  units  in  the 
multiplier. 

But  to  subtract  — 4  from  zero  once,  gives  +4,  (Art.  8,) 
and  to  subtract  it  3  times  as  the  — 3  indicates,  gives  +12. 
That  is,  minus  multiplied  into  minus,  gives  plus. 

This  principle  is  so  important  that  we  give  another  mode 
of  illustrating  it,  making  use  of  the  following  example.* 

Required  the  product  of  a — h  by  a — c. 

Here  a — h  must  be  repeated  a — c  times. 

*  There  is  also  another  method  of  showing  that  minus  multiplied 
into  minus,  must  give  plus  ;  and  it  rests  on  the  principle  that  a  times 
0  gives  0  for  a  product,  or  0  times  any  quantity  must  give  0.  In  short, 
the  product  of  two  factors  must  be  zero,  if  either  one  of  them  is  zero. 

Suppose  we  multiply     .     .       a — a 

By b 

The  product  is      ....     ah— ah 

Here  a — a  is  in  value  0.     So  in  the  product  ah — ah  is  0,  as  it  should 
be,  and  the  whole  subject  is,  thus  far,  very  clear. 
Now  supppose  we  take 
And  multiply  by       ...    — h 


The  product  is      .     .     .      —HLh-\-ah 

The  first  part  of  the  product  is  clearly  — ah,  and  the  whole  must  be 
zero  ;  therefore  we  must  take  the  second  part,  -\-ah,  to  destroy  the  first, 
that  is,  — h  multiplied  by  ^—a,  gives  -\-ah. 

The  objection  to  this  method  is,  that  the  reasoning  at  the  last  point  is 
rather  m^clianical  than  intellectual  ;  we  are  forced  to  take  ah  as  plus  to 
make  a  definite  sum,  giving  no  decided  metaphysical  reason  that  it 
must  be  so. 


MULTIPLICATION.  39 

If  we  take  a — 5,  a  times,  we  shall  have  too  large  a  product, 
as  the  multiplier  a  is  to  be  diminished  by  c. 
That  is  a—h 

Multiplied  by      a 

Gives  .  .  .  aa — ahy  which  is  too  great  by  a — 5  repeated 
c  times,  or  by  ac — c5,  which  must  be  subtracted  from  the 
former  product ;  but  to  subtract  we  change  signs,  (Art.  5,) 
therefore  the  true  product  must  be  aa — ah — ac-{-cb. 

That  is,  the  product  of  minus  h,  by  minus  c,  gives  plus  be, 
and,  in  general,  minus  multiplied  hy  minus  gives  plus. 

But  plus  quantities  multiplied  by  plus  give  plus,  and  minus 
by  plus,  or  plus  by  minus,  give  minus;  therefore  we  may 
say,  in  short. 

Thai  quantities  affected  hy  like  signs,  when  multiplied  together, 
give  plus,  and  when  affected  hy  unlike  signs,  give  minus. 

(Art.  14.)  The  product  of  a  into  b  can  only  be  expressed 
by  ah  or  ha.  The  product  of  a,  b,  c,  d,  <fec.,  is  ahcd;  but  if 
b,  c,  and  d  are  each  equal  to  a,  the  product  would  be  aaaa. 

The  product  of  aa  into  aaa  is  aaaaa;  but  for  the  sake  of 
brevity  and  convenience,  in  place  of  writing  aaa,  we  write  a^. 
The  figure  on  the  right  of  the  letter  shows  how  many  times 
the  letter  is  taken  as  a  factor,  and  is  called  an  exponent.  The 
product  of  a?  into  a*  is  a  repeated  3  times  as  a  factor,  and  4 
times  as  a  factor,  in  all  7  times ;  that  is,  write  the  letter  and 
add  the  exponents. 

EXAMPLES. 

What  is  the  product  of  a^  by  a^  ? 
What  is  the  product  of  rr"*  by  x^  ? 
What  is  the  product  of  3/^  by  y*  by  j^^ 
What  is  the  product  of  a"  by  a^  ? 
What  is  the  product  of  6V  by  bx  ? 
What  is  the  product  ©f  <w  by  oc^  by  aV  ? 


Ans. 

a\ 

Ans. 

x'\ 

9 

Ans. 

f\ 

Ans. 

QU+m^ 

Ans. 

b'x\ 

.¥? 

Ans. 

a'<?. 

40  ELEMENTARY   ALGEBRA. 

What  is  the  product  of  ar^  by  a:^  by  ar*  ?  .     .  Arts.         sP. 

What  is  the  product  of  a^  by  x^  by  a;"*  ?       .  Ans.  a^+^  ^^. 

What  is  the  product  of  S:t^  hj2x^hj2'>     .  Ans.      12ar\ 
Find  the  product  in  each  of  the  following  examples : 

4ac  9a^c  — 3xy  — 2xi/ 

— 3(ib  — 4ay  +9a:y  — 6xi/ 


Product 


-laij  210a;y  40rt  —21p 

3x7/  — Say  20pq  —  3r 


*  (Art.  15.)  The  preceding  examples  are  sufficient  to  illus- 
trate the  multiplication  of  simple  factors — we  now  proceed  to 

Case  2 .  When  one  of  the  factors  is  a  compound  quantity, 
we  have  the  following 

Rule  . — Multiply  every  term  of  the  multijolicand,  or  com- 
pound quardity,  separately,  hy  the  multiplier,  a^  in  the  former 
case;  placing  the  products  one  after  another,  with  the  p)foper 
signs;  and  the  result  will  he  the  whole  product  required. 

The  reason  of  this  rule  is  obvious  from  Case  1 . 

EXAMPLES. 

(1)  (2)  (3) 

5a — 3c  3ac — 4^  2a* — 3c+5 

2a  3a  he 


10a*— 6ac  2ah—\^ah  ^a^bc—3be-\-bbc 

(4)  (6)  (6) 

l,2ar — 2ac  26c — Ih  Ax — h-^3ah 

Aa  — 2a  2aJ 


MULTIPLICATION.                                  41 

(7)                                (8) 

3(?-^x                        10^—3/ 

(9) 

Axy                          — 4x^ 

^ax" 

10.  Multiply  36— 2c  by  56.  .     . 

.     .     Ans.   156^— 106c. 

11.  Multiply  Axy—d  by  Qx.       . 

.     .     Ans.  24ar''y — 3'ix. 

12.  Multiply  a^— 2a;-M  by  A:t^. 

Ans.   4a2a;2_8^3_|_4^2^ 

13.  Multiply  llaW— 13^y  by  2,ax. 

Ans.  SSa'bc'x—S9ax^y. 

14.  Multiply  42c2— 1  by  —4. 

^725.  —  168c2+4. 

15.  Multiply  —2>Q)a^ha^y-\-\2,  by 

16.  Multiply  26— 7a— 3  by  4a6 

—5a\        ... 

Ans.  -{-150a^bx^y~65aK 

Ans.  Qab^—2Qa^b—'i2ab. 

17.  Multiply  a+36— 2c  by  — 3a6. 

Ans.  — 3a26— 9a6'+6a6c. 

18.  Multiply   13a'— 62c  by  —4c.         Ans.  —52ah+4b'c\ 

19.  Multiply  ISary- 36  by  —25^:2^ 

Ans.  —325x^y—75bx\ 
Case  3 .      When  both  the  factors  are  comjjound  quantiiies, 
we  have  the  following 

Rule  . — Multiply"  every  term  of  the  multiplicand  by  evei-y 
term  of  the  multiplier,  separately ;  setting  down  the  products  one 
after  or  under  another,  with  their  proper  signs ;  and  add  the 
several  lines  of  products  all  together  for  the  whole  product 
required. 


EXAMPLES. 

(1)                                     (2) 

Multiply     .     .  2a-}-36                  Qa^y   —2z 
By    .     .     .     .     a-f  ^                   Sax    —bd 

Product  by  a     2a^-\-Sab 
Product  by  6               2a6-|-362 

Entire  product   2a'-|-5a6-f-36' 
4 

1  ^aa^y—Qaxz—SQ^dxy^  1  Qdz 

42 


ELEMENTARY  ALGEBRA. 


3.  Multiply  a-\-h-{-c  by  x-{-y-{-x,  that  is,  repeat  a-\-h-\-c,  x 
times,  then  y  times,  then  z  times,  and  the  operation  stands 
thus: 

a-\-h-]rc 
x-\-y-\-z 


Product  by  x 
Product  by  y 
Product  by  s 

Entire  product 

4.  Multiply 
By 
Partial  product 
2d  partial  product 

Whole  product 
5.  Multiply  3a2 


ax-\-hx-\-cx 
ay^hyArcy 

az-\-bz-\-cz 


ax-\-bx-\-cx-\-ay-\-by-{-cy-\raz-{-bz-{-cz. 


2a^'-{-xy — 2y^ 
3x  — 3y 


ex^-j-Sx'y—Gxf 
—Sx'y—SxfJf-ef 

ex'—^x^y—dxy'^+Gf 


■2ab—b'^  by  2a— 45. 

Frod.  6a^—16a^b-{-6ab'^+4b\ 

6.  Multiply  x^ — xy-\-'f  by  a;-hy.      .     .     .  Prod.  aP-\-y^. 

7.  Multiply  3a-l-4c  by  2a — 5c.        Arts.  Ga^ — 7ac — 20c^. 

8.  Multiply  a^-j-ay — y^  by  a — y.      Ans.       a? — 2ay'^-\-y^. 

9.  Multiply  a^-\-ay-\-'f  by  a — y. 

10.  Multiply  a^ — ay+y^  by  a-\-y. 

11.  Multiply  a'^+aV+ay^+y^  by  a — y. 

12.  Multiply  f — y-[-\  by  y+1. 

13.  Multiply  a^-Vf  by  ar^— /.      . 

14.  Multiply  a2_3a_|_8  by  a+3.     , 

15.  Multiply  5^4-5V-fa;^   by  i^— a:^^ 

16.  Multiply  rt'^+i"*  by  a+Z*. 

17.  Multiply  a;^+a;'*+^  by  or — 1.      .     Ans.  a? — x^. 

18.  Multiply  771+71  by  ^m — 9w.    .     .     Ans.       9m^ — 9n^. 


Ans.  a^ — y^. 
Ans.  a'+y*. 
Ans.  a^ — y*. 
Ans.  y^-j- 1 . 
An^.  x^ — y'^. 


Ans. 


5-1-24. 


A71S.  b^ — x^.. 


MULTIPLICATION.  43 


19.  Multiply  2/2_20  by  t/'-^+SO. 

20.  Multiply    a+5  by  a-\-b.     . 

21.  Multiply    rr+y  by  x-\-y.    . 

22.  Multiply    a — h  by  a — h.     . 

23.  Multiply    x — y  by  x — y.    . 


-4n5.  y^ — 400 
Ans.  a2+2a5+62 
Arts.  ^Ar^xy-T'if- 
Ans.  c? — ^ah-\-W- 
Ans.  a? — 9,xy-\-'t^ 


(Art.  16.)  When  a  number  is  multiplied  by  itself,  the 
product  is  called  its  square,  the  square  of  one  of  the  fac- 
tors, and  by  inspecting  the  last  four  examples,  we  perceive 
that  the  square  of  any  binomial  quantity,  (that  is,  the  square 
of  any  two  terms  connected  together  by  the  sign  plus  or 
minus),  the  result  must  be  the  squares  of  the  two  parts, 

AND  twice  the  PRODUCT  OF  THE  TWO  PARTS. 

N.  B.  The  product  of  the  two  parts  will  be  plus  or  minus, 
according  to  the  sign  between  the  terms  of  the  binomial. 
By  this  summary  process  perform  the  following  examples : 

1.  Square  (3a+5)  or  multiply  this  quantity,  by  itself  con- 
sidered as  a  numeral  quantity.*  Ans.  ^d?-\-Qah-\-l?. 

2.  Square  2a; — y.  Ans.  4x'^ — 4xy+y^. 
We  write  the  product,  in  the  second  place,  in  the  answer, 

because  it  naturally  falls  there  when  the  multiplication   is 
formally  made  ;  but  this  is  not  essential. 

Write  out  the  following  squares  as  indicated  by  ike  exponent. 

(a--3cy=a'—6ac-\-9c^     . 
(3a__c)2=9a2_l-6ac+c2 

(2a;4-3y)==4a;2-{- 1 2:ry+92/2 
(20x-]-yy= 400a;2-f  40xy-]-7/ 

*  We  make  this  last  remark  because  things,  arithmetically,  cannot  ])e 
multiplied  by  things.  For  instance,  dollars  cannot  be  multiplied  by 
dollars,  &c.  In  fact,  every  multiplier  is  always  a  number  ;  and  when 
we  demand  the  square  or  any  other  power  of  a  quantity,  it  always 
means  the  power  of  its  numeral  value  considered  abstractly. 


44  ELEMENTARY  ALGEBRA. 

(Art.  17.)  The  product  of  the  sum  and  difference  of 
two  quantities  is  equal  to  the  difference  of  their  squares, 
as  will  be  seen  by  inspecting  the  followiny  products  : 

The  first  example  should  be  multiplied  in  full  to  establish 
the  principle. 

What  is  the  product  of  {a-\-b)  by  (a — h)  ? 

Ans.  a? — i^. 
What  is  the  product  of  2m4-2w  by  2m — 2«  ? 

Ans.  4m? — 4n^. 

What  is  the  product  of  x-\-i/  by  x — ^y?  Ans,  x^ — y^. 
What  is  the  product  of  %x-\-2>y  by  3a; — 3y  ? 

Ans.  92;2— 92/2. 
What  is  the  product  of  la-\-b  by  la — b  ? 

Ans.  49a2— 52. 
What  is  the  product  of  1  +  lOa  by  1— 10a? 

Ans.  1—1  OOa^. 

Observation. — By  attention  to  principles  much  labor  may  be 
saved  in  the  common  operations  of  Algebra.  For  instance, 
if  the  product  of  three  equal  binomial  factors  were  required, 
as  (x'\-^)[x-\-2>){x-\-^),  we  may  first  write  out  the  product 
of  two  of  those  factors. by  (Art.  16);  then  multiply  that 
product  by  the  other  factor. 

Thus,     ....     x''-\-e>x-\-9 
a: +3 

?+^6?+9a: 

3a:2_j-i8a:-f-27 

Product       .     .     .    "^9^+27^+27 

If  the  product  of  the  four  factors,  {x — A){x — 5)(a:-}-4) 
(^x-\-b),  were  required,  we  would  take  the  product  of  ihe  first 
and  third  factors,  then  of  the  second  and  fourth,  by  (Art. 
17),  then  the  product  of  those  two  products  would  be  the 
final  product  required. 


DIVISION. 


45 


Thus,  the  required  product  is  the  product  of  (x^ — 16)  by 
(a;2_25)  =2:^—4  la;2-l-400. 

What  is  the  product  of  (a-\-c)(a-{-d)(a — c)(a — d)  ? 

Ans.  a^ — a^c^ — a^d^-{-cW. 


DIVISION. 

(Art.  18.)  Division  is  the  converse  of  multiplication,  the 
product  being  called  a  dividend,  and  one  of  the  factors  a 
divisor.  If  a  multiplied  by  b  give  the  product  ab,  then  ab 
divided  by  a  must  give  b  for  a  quotient,  and  if  divided  by  5, 
give  a.  In  short,  if  one  simple  quantity  is  to  be  divided  by 
another  simple  quantity,  the  quotient  must  be  found,  by  in- 
sjpection,  as  in  division  of  numbers. 


EXAMPLES. 

1.  Divide  1 6a5  by  4a *Ans. 

4b. 

2.  Divide  2\acd  by  7c.     . 

,     .     Ans. 

Sad. 

3.  Divide  ab^c  by  ac. 

.     Ans. 

b\ 

4.  Divide  6abc  by,  2c. 

.     Ans. 

3ab. 

6.  Divide  cui^  by  ax^. 

,     Ans. 

X. 

6.  Divide  Smx^  by  mx.    . 

.     Ans. 

Z'.'^. 

7.  Divide  2\0c^b  by  7cb.      . 

,     Ans. 

SOc\ 

8.  Divide  42a^  hy  xy.      ,     , 

.     Ans. 

42. 

9.  Divide  3xy  by  ax. 

,     Ans. 

^1. 
a 

*  The  term  quotient  would  be  more  exact  and  technical  here  ;  but, 
U.'.  results  hereafter,  we  shall  invariably  use  the  term  Ans.,  as  more  brief 
and  elegant,  and  it  is  equally  well  understood. 


46 


ELEMENTARY  ALGEBRA. 


Remark. — In  this  last  example  we  cast  out  the  equal 
factor  X  from  both  the  dividend  and  divisor,  and  set  the  other 
factor  a  of  the  divisor  under  the  dividend  as  a  denominator. 

(Art.  19.)  When  the  dividend  and  divisor  have  no  factors 
in  common,  we  can  only  indicate  the  division  by  setting  the 
divisor  under  the  dividend  for  a  denominator,  as  in  the  follow- 
ing example : 


Divide  Zahc  by  2xy, 

.     .     .     Ans, 

Sabc 
2xy 

Divide  Aaxy  by  Zay, 

.     ,     .     Ans. 

4x 

Divide  2)Qaby  by  4a5y.     . 

.     .     .     Ans. 

9. 

Divide  9.1  aby  by  Wabx. 

.     .     .     Ans. 

27y 
llx 

Divide  19J}''x  by  2,abx. 

/    * f\r\    \     T_      J.!.  -     J 

.     .     .     Ans. 

9b 
a 

(Art.  20.)  In  the  preceding  examples  no  signs  were  ex- 
pressed, and,  of  course,  every  term  and  every  factor  is  under- 
stood to  be  positive ;  but  as  algebraic  quantities  may  have 
negative  signs,  and  unlike  signs,  we  must  investigate  and 
decide  upon  the  sign  to  prefix  to  the  quotient.  This  can  be 
done  by  merely  observing  what  sign  must  be  put  to  the  quo- 
tient so  that  the  product  of  the  divisor  and  quotient  will  give 
the  same  sign  as  in  the  dividend,  according  to  the  principles 
laid  down  in  multiplication,  (Art.  13). 

For  example,  divide  — 9y  by  3y,  the  quotient  must  be  — 3 ; 
so  that  3y  multiplied  by  — 3  will  give  — 9y,  the  dividend. 

Divide  — Py  by  — 3y,  the  quotient  must  be  ~\-3. 
Divide  4-9y  by  — 3y,  the  quotient  must  be  — 3. 

From  these  examples  we  draw  the  following  rule  for  the 
signs : 

Rule. —  When  the  dividend  and  divisor  have  like  signs, 
hoik  -j-  or  both  — ,  then  the  quotient  must  be  plus. 


DIVISION.  47 

When  the  dividend  and  divisor  have  unlike  signs,  the  quotient 
must  he  minus.* 

EXAMPLES, 

Divide  — 2 lac  by  — la Ans.    +3c. 

Divide  — \9,xy  by  4-3y Ans.    — 4a:. 

Divide  I'^abc     by  — 8c Ans.  — 9aZ>. 

Divide  \Umn   by  +8ac Ans.  —^. 


•  *  N  o  T  E . — We  address  this  note  to  those  only  who  are  fond 
of  the  Tnetaphysique  of  science.  Division,  considered  in  its  most 
elementary  sense,  is  not  merely  the  converse  of  multiplication ; 
it  is  a  short  process  of  finding  how  many  times  one  quantity 
can  be  subtracted  from  another  of  the  same  kind.  When  the 
suhraction  is  possible,  and  diminishes  the  humeral  value  of  the 
minuend,  and  brings  it  nearer  to  zero,  the  operation  is  real  and 
must  be  marked  plixs.  When  the  subtraction  is  not  possible 
without  going  farther  from  zero,  we  must  take  the  converse  oper- 
ation, and  the  converse  operation  we  must  mark  minus. 

Thus,  divide  18a  by  6a.  Here,  it  is  proposed  to  find  how  many 
times  6a  can  be  subtracted  from  18a  ;  and  as  we  can  actually  sub- 
tract it  3  times,  the  quotient  must  be  -|-3. 

Divide  — 18a  by  — 6a.  Here,  again,  the  subtraction  can  actu- 
ally  be  performed,  and  the  number  of  times  is  3,  and,  of  course, 
the  quotient  is  -|-3. 

Divide  — 18a  by  6a.  Here,  subtraction  will  not  reduce  the 
dividend  to  zero  ;  but  addition  will,  and  must  be  performed  3 
times  ;  but  the  operation  is  the  converse  of  the  one  proposed, 
and  therefore  must  be  marked  by  the  converse  sign  to  plus,  that 
is  —3. 

Again,  divide  18a  by  — 6a.  Here,  if  we  sub,  — Qa  it  will 
not  reduce  18a ;  but  the  converse  operation  will,  and  therefore 
the  quotient  must  be  minus,  that  is,  — 3. 

Now  let  us  inspect  the  common  operation  of  division,  by  the 
help  of  the  following  example  :  Divide  24  by  8.  Let  the  oper- 
ation stand  thus : 


48  ELEMENTARY   ALGEBRA. 

(Art.  21.)  The  product  of  a^  into  a^  is  a^  (Art.  14),  that 
is,  in  multiplication  we  add  the  exponents ;  and  as  division  is 
the  converse  of  multiplication,  to  divide  powers  of  the  same 
letter,  we  must  subtract  the  expone7it  of  the  divisor  from  that  of 
the  dividend. 

1.  Divide  2a^  by  a'' Ans.  2a 

2.  Divide  — a'  by  a^ Ans.  — c 

.3.  Divide  \6x^  hj  4x Ans.  4a^. 

4.  Divide  IBaxy^  by  — Say Ans.     — 5xi/^ 

5.  Divide  63a"*  by  7a" Ans.      9a'"-'' 

6.  Divide  I2ax^  by  — 3aa; Ans.  — 4x'^~^ 

7.  Divide  28ay  by  4ac2^ Ans.         1^ 

c 

Divisor.        Divi.        Quotient. 
8)24(3 
The  product  of  the  divisor  and  quotient,  in  all  cases,  equals 
the  dividend.     Let  d  represent  any  divisor,  D  any  dividend,  and  q 
the  corresponding  quotient,  then 

dq=D 

Or d^jE^ 

9 

In  the  above  numeral  example,  let  us  suppose  the  divisor  8  to 
be  — 8,  and  the  quotient  — 3.  Then  the  dividend  must  be  the 
product  of  ( — 8)X( — 3);  but  suppose  that  we  do  not  know 
whether  this  is  plus  or  minus,  we  will  therefore  represent  it  by  D. 

Then (— 8)(— 3)=Z> 

By  dividing  both  members  by  either  factor,  as  ( — 3),  we  have 

-8=A  or  J?_=  -8 
—3         —3 

Here  D  cannot  be  minus,  for  minus  divided  by  minus  must  give 

plus  in  the  quotient ;  (as  we  have  just  determined  in  this  note), 

but  the  quotient  is  actually  ( — 8),  therefore  D  must  be  plus.     That 

is,  the  product  of  minus  into  minus  gives  plus  ;  corresponding  to 

(Art.  13). 


DIVISION.  40 

8.  Divide  — ISa^x  by  — 6ax.       .     .     .     Ans.  3a*, 

9.  Divide  ^acdxif-  by  SLadxif-.       .     .     .     Atis,  3c. 

10.  Divide  ^b{a—xf  by  \b{a—xf.       .     Ans.  3(a— a;). 

In  this  last  example  consider  {a — x)  as  one  quantity. 

11.  Divide  45y^  by  \tAf- Ans.  3y. 

Examples  10  and  11  are  exactly  alike,  if  we  conceive 
(a — x)  equal  to  y. 

12.  Divide  12aV  by  — ^o^x Ans.       — 4a?. 

13.  Divide  \bmf-  by  Say Ans.       — by. 

14.  Divide  — -18aa:^y  by  — 8ac2.     .     .     .     Ans.         ^, 

15.  Divide  7a^b  by  2l;^b\    .     . 

16.  Divide  —5a^a^  by  — 7aV. 

17.  Divide  117a'bY  by  TSa^^c^ 

18.  Divide  ^  hj  ^ Ans. 

19.  Divide  (x—^f  by  (a;— y)'.      .     .     .    Ans.  (x-^yf. 
Observe,  that  example  18  and  19  are  essentially  alike. 

20.  Divide  {a-\-by  by  (a+b).        .     .     .     Ans.  (a+b)*. 

21.  Divide  (a—cy'*  by  (a — c)».     .     .     Ans.  (a — c)'»-«. 

To  perform  example  21  we  adhere  to  the  principle  of  per- 
forming 18,  19,  and  20. 

(Art.  22.)  In  the  process  of  division,  exponents  may  be- 
come negative,  and  it  is  the  object  of  this  article  to  explain 
their  import. 

To  explain  this,  take  a*  and  divide  successively  by  a,  form- 
ing the  following  series  of  quotients : 


Ans 

7a^b 
Slla'b'' 

1 
^?iab 

• 

Ans. 
Ans. 

5 

70,^' 
36' 
2c' 

, 

Ans. 

'^. 

5 


cf,    a,     1,     _,     _^,        ,     dfc. 


1             -2, 

1        -' 

_=a    ' 

a"- 

a« 

50  ELEMENTARY  ALGEBRA. 

Divide  a'*  successively  by  a  again,  rigidly  adhering  to  the 
principle  that  to  divide  any  power  of  a  by  a,  the  exponent 
becomes  one  less,  and  we  have 

a',     a',     aS     a",     a~S     a~^     a~^,     &c. 

Now  these  quotients  must  be  equal,  that  is,  a?  in  one  series 

equals  a'  in  another,  and 

1       ~* 
a^=c^y     a=a},     l=a°,     -=a    , 
a 

Another  illustration.     We  divide  exponential  quantities  by 

subtracting  the  exponent  of  the  divisor  from  the  exponent  of 

the  dividend.     Thus,  d^  divided  by  a^  gives  a  quotient  of 

c^^=a^.     aP  divided  by  a''=  a^~'=a~^.     We  can  also  divide 

by  taking  the  dividend  for  a  numerator  and  the  divisor  for  a 

c^      1  1 

denominator,  thus,  _=_,  therefore,  _=a~^  (Axiom  7). 

From  this  we  learn,  that  exponential  terms  may  be  changed 
frum  a  numerator  to  a  denominator,  and  the  reverse ^by  changing 
the  signs  of  the  exponents. 

Thus,  %==a^^    -^::=-l3    ^=:.- 

Divide  cc'hc  by  c^h^c'^ Ans.  a'^lr^c. 

Observe,  that  to  divide  is  to  subtract  the  exponents. 

Divide  aV  by  aVy^.      .     .     .     Ans.  or  a~^or^y~^. 

a^xy"^ 

3  3 

Divide  2)ay'^  by  bc^x^y"^.  .     .     .     Ans.     — ^—  or  -a~^x~^, 

(Art.  23.)  When  the  dividend  is  a  compound  quantity, 
and  the  divisor  a  simple  (or  single)  quantity,  we  have  the  fol- 
lowing rule,  the  reason  of  which  will  be  obvious  if  the  pre- 
ceding part  of  division  has  been  comprehended. 

Rule. — Divide  each  lei'm  of  the  dividend  by  the  divisor, 
and  the  several  results  connected  together  by  their  proper  signs 
wUl  be  the  quotient  sought. 


DIVISION  51 

EXAMPLES. 

1.  Divide  \5ab — l^ax  hj  3a Ans.  5b — 4x, 

2.  Divide  — 25a^x-\-15aa^  by  — Box.  .     .  Ans.  5a — 3x. 

3.  Divide  10ab-\'15ac  hj  5a Ans.    2^-1- 3c. 

4.  Divide  SOax — 54x  by  6a: Ans.    5a — 9. 

5.  Divide  8a;3-j-12a:2  |jy  4^ ^^^_     2^:4-3, 

6.  Divide  3bcd-{-\2bcx — %h  by  35c.    .    Ans.  d-\-Ax — 3h, 

7.  Divide  lax-\-3ay — 75c?  by  — lad, 

S  Am.  — ^_?2^-{-i 

d     Id    a 

8.  Divide  3ao^-\-Qx^-\-3ax — 15a;  by  3a;. 

Ans.  a.7^-{'2x-\-a — 5. 

9.  Divide  3a5c+12a5a; — 3a^5  by  3a5.        Ans.  c-\-4x — a. 

10.  Divide  25a^bx — 15a^ca;^-t-5a5c  by  — 5ax. 

Ans.  — 5a5-|-3aca; — bcx~^. 

11.  Divide  20a5^+15a62+10a5+5a  by  5a. 

Ans.  453+352+25-1-1. 

(Art.  24.)  We  now  come  to  the  last  and  most  important 
operation  in  division,  the  division  of  one  compound  quantity 
by  another  compound  quantity. 

The  dividend  may  be  considered  a  product  of  the  divisor 
into  the  yet  unknown  factor,  the  quotient ;  and  the  highest 
power  of  any  letter  in  the  product,  or  the  now  called  divi- 
dend, must  be  conceived  to  have  been  formed  by  the  highest 
power  of  the  same  letter  in  the  divisor  into  the  highest  power 
of  that  letter  in  the  quotient.  Therefore,  both  the  divisor  and 
the  dividend  must  be  arranged  according  to  the  regular  powers 
of  some  letter. 

After  this,  the  truth  of  the  following  rule  will  become 
obvious  by  its  ^eat  similarity  to  division  in  numbers. 

Rule  . — Divide  the  first  term  of  the  dividend  by  the  first 
term  of  the  divisor,  and  set  the  result  in  the  quotient.* 

*  Divide  the  first  term  of  the  dividend  and  of  the  remainders  by  the 
first  term  of  the  divi8<>r  ;  b©  not  troubled  about  other  term*. 


52  ELEMENTARY  ALGEBRA. 

Multiply  the  whole  divisor  hy  the  quotient  thus  found,  and 
subtract  the  product  from  the  dividetid. 

The  remainder  will  form  a  new  dividend,  with  which  proceed 
as  before,  till  the  first  term  of  the  divisor  is  no  longer  contained 
in  the  first  term  of  the  remainder. 

The  divisor  and  remainder,  if  there  he  a  remainder,  are  then 
to  he  written  in  the  form  of  a  fraction,  as  in  division  of  numbers. 


EXAMPLES. 

Divide  a^-{-2ah+h^  by  a-\-b. 

Here,  a  is  the  leading  letter,  standing  first  in  both  dividend 
and  divisor :  hence  no  change  of  place  is  necessary. 


OPERATION. 

a-\-h)a''-\-2ab+h\a-\-h 
a^-jr  ah 

ah-\-l>' 
ab-\-h^ 

That  the  pupil  may  perceive  the  close  connection  between 
multiplication  and  division,  we 


Multiply        a^-\-2ab-\-4b^ 
By  2a^—2ah-\-b'^ 


2a'-\-4a:'h-{-Qa^h^  (1) 

+d'b^-h2ab^-\-4b^         (3) 
Prod,  is         2a'-{-2a''h-\-5d'h^'-6ab'-{-4h* 

Now  take  this  product  for  a  dividend,  and  one  of  the  fac- 
tors, (c^-{-2ah'{-4h^),  for  a  divisor,  and  of  course  the  other 
factor,  (2a^ — 2ah-\-h^),  will  be  the  quotient,  and  the  operation 
will  stand  thus  r 


DIVISION.  53 

(1) 


(2) 

(3) 


2a'-{-4a'b+8a'b^ 

—2a'b—3a^b^- 
—2a^b—4a'b'- 

-6ab^ 
-Qab^ 

a'b'+2ab'+4b' 
a'b^-\-2ab'-{-4b* 

The  several  partial  products  which  make  up  the  dividend, 
'^  and  marked  (1),  (2),  (3),  are  again  found  in  the  operation  of 
division,  and  there  marked  (1),  (2),  (3),  the  same  as  in 
Arithmetic. 

Some  operators  put  the  divisor  on  the  right*  of  the  divi- 
dend, as  in  the  following  example  : 

Divide  a^ — b^  by  a — b. 

a — h 


-b'     )   a—b 

a'b—ab^ 
ab'^—b^ 
ab^—b' 


b-\^b\  Quo. 


GENERAL    EXAMPLES. 

1.  Divide  a^+2aa;+ar*  by  a-^-x.  Ans.  a-\-x. 

2.  Divide  a? — 3a?y-\-3ay'^ — y^  by  a — y. 

m  Ans.  a? — 2ay-l-3/'« 

3.  Divide  9,Aa''b—nahb'^—Qab  by  — 6a5. 

Ans.  — 4a-\-2a?cb-\-'[. 

4.  Divide  a?-[-ba?b^-5ab'^-\-W  by  a+5. 

Ans.  a?-{-4ab-\-b'^. 

*  N  o  T  E . — This  is  in  imitation  of  the  French,  and  being  a  mere  matter 
of  taste,  involving  no  principle,  we  have  no  right  to  find  fault  with  those 
who  adopt  it ;  and  others  must  not  complain  of  us  because  we  prefer 
tlie  English  custom. 


54  ELEMENTARY  ALGEBRA. 

5.  Divide  a^+^a'^b-\-2aL^+P  by  a^+ai+i^.  Ans.  a+L. 
0.  Divide  ic^— 9a;2+27a;— 27  by  ar— 3.  Ans.  x^—6x-{-d. 
7.  Divide  6a;*— 96  by  6x— 12.  Ans.  a^+2a;2+4a;+8. 

(Art.  25.)  When  a  factor  appears  in  every  term  of  both 
dividend  and  divisor,  it  may  be  cast  out  of  every  term  with- 
out affecting  the  quotient ;  thus,  in  the  last  example,  the  fac- 
tor 6  may  be  cast  out  by  division ;  and  x* — 16  divided  by 
X — 2  will  give  the  same  quotient  as  before. 

8.  Divide  ea-'+Qa^— 15a  by  3a^—3a. 

Ans.  2a2+2a4-5. 
(Observe  Art.  25). 

9.  Divide  252r^— ar^— 2a;2— 8.^  by  5x^—4x. 

Ans.  5a^+4a:2+3a:+2. 

10.  Divide  ISa^—Sb^  by  6a-f  45.  Ans.  3a— 2b. 

11.  Divide  22;^— 19a;2+26a;— 16  by  x—S. 

Ans.  2x^ — 3a;4-2. 

12.  Divide  2/'+l  by  y-\-l.  Ans.  y"^ — y^-\-y^ — y+1. 

13.  Divide  /—I  by  y—\.     Ans.  f-\-y'^^-f-\-y^-Vy-\-\. 

14.  Divide  ar^ — a^  by  x — a.  Ans.  x-\-a. 

15.  Divide  Qa?—3a~b—2a-^b  by  3a2— 1.        Ans.  2a— h. 

16.  Divide  /— 3yV+3yV— a;*'  by  f—3y'^x-\-3ijx'^—x\ 

Ans.  2/'*+3y2a;4-3ya;2+ar3. 

17.  Divide  64a*6«— 25a=Z»«  by  ^a%^-\-bah\ 

Ans.  Sa-b^-Sab^ 

18.  Divide  2a'* — 2a;*  by  a—x. 

Ans.  2a^-l-2a=^a;4-2aar'4-2ar». 

19.  Divide  (a — xy  by  (a — xy.  Ans.  (a — xf. 

20.  Divide  a^— 3a2a:4-3aa;2— a;^  by  a—x. 

Ans.  a^ — 2aa;+a-^. 

21.  Divide  a'^+1  by  a+l.  Ans.  a*— a^'+a^— a-f-l. 

22.  Divide  b^—\  by  b—\.        Ans.  lf'^b''^b^-\-b--\-b-\-\. 


DIVISION.  55 

23.  Divide  4Sa^—92a^x—40ax^-\-100x^  by  3a— 5x. 

Ans.   lCia^—4ax—20x^. 

24.  Divide  4d'—9d^-{-6d—l  by  ^d'^+Zd—l. 

Ans.  2^2— 3c^4-l. 

25.  Divide  10ab+15ac  by  2b-\-3c. 

26.  Divide  SOao;— 54a;  by  5a — 9. 

27.  Divide  Sx^-^nx"  by  2a;-f3. 

28.  Divide  — 25a^x-{-15ax^  by  5a — 3x. 

Observe  that  these  last  four  examples  are  the  same  as  some 
m  (Art.  23.) 

If  more  examples  are  desired  for  practice,  the  examples  in 
multiplication  may  be  taken.  The  product  or  answer  may  be 
taken  for  a  dividend,  and  either  one  of  the  factors  for  a  divi- 
sor ;  the  other  will  be  a  quotient. 

Also,  the  examples  in  division  may  be  changed  to  examples 
in  multiplication ;  and  these  changes  will  serve  to  impress  on 
the  mind  of  the  pupil  the  close  connection  between  these  two 
operations. 

(Art.  26.)  The  operation  of  division  is  the  art  of  finding 
one  of  two  factors  of  a  product,  when  the  product  itself 
and  one  factor  is  given.  When  the  product  only  is  pre- 
sented,  and  its  factors  required,  the  operation  is  properly 

called 

FACTORING. 

Factors  of  a  number  are  such  numbers  as  may  be  multi- 
plied together  to  produce  the  number  ;  and  factors  of  an  al- 
gebraic expression  are  such  quantities  as  being  multiplied 
together  will  produce  the  expression.  Thus,  2  and  3  are 
the  factors  of  6,  because  2X3=6,  and  3,  a,  and  c,  are  fac- 
tors of  3ac,  because  by  their  multiplication  they  form  that 
product. 

But  some  numbers  /taw  wo/adors,  (except!  and  the  number 


56  ELEMENTARY  ALGEBRA. 

these  should  not  be  considered  factors),  and  such  numbers 
are  called  prime  numbers. 

Also,  some  algebraic  expressions  have  no  factors,  and  such 
expressions  are  called  prime  quantities.  Thus,  5a-^c  is  a 
prime  quantity. 

The  following  is  a  list  of  the  prime  numbers  up  to  100 : 

1,  2,  3,  5,  7,   11,  13,  17,  19,  23,  29,  31,  37,  41,  43,  47, 
53,  59,  61,  67,  71,  73,  79,  83,  89,  97. 

All  the  other  intermediate  numbers  are  composite  numbers. 
Any  number  whatever  is  either  a  prime  number,  or  composed 
of  the  product  of  2^rime  factors. 

Knowing  this  fact  enables  us  to  decompose  any  number  into 
its  prime  factors  by  the  following  rule  : 

Rule  . — Divide  the  given  number  by  any  prime  number  that 
will  divide  it  without  a  remainder,  and  divide  that  quotient  again 
by  any  prime  number  that  will  exactly  divide  it,  and  so  continue 
until  the  last  quotient  is  a  prime  number.  The  divisors  and 
last  quotient  are  the  factors  required. 

N.  B.  Use  the  smallest  prime  divisors  first. 

EXAMPLES. 

1.  Required  the  factors  composing  102.       Ans.  2,  3,  17. 

OPERATION. 

2)102 

"~17 

2.  Find  the  prime  factors  in  the  number  112. 

Ans.  2,  2,  2,  2,  7. 

3.  Find  the  prime  factors  in  the  number  126. 

Ans.  2,  3,  3,  7. 

4.  Find  the  prime  factors  in  the  number  12769. 

Ans.    113,  113. 


FACTORING.  57 

5.  Find  the  prime  factors  in  the  number  1156. 

Ans.    2,  2,  17,  17. 

6.  Find  the  prime  factors  in  the  number  1014. 

Ans.  2,  3,  13,  13. 

(Art.  27.)  As  the  combination  of  numbers  is  endless,  it  is 
impossible  to  give  any  definite  rule  that  will  decide  in  each 
and  every  case  -whether  a  number  is  a  prime  or  a  composite 
number ;  and  as  the  practical  utility  of  factoring  is  limited,  it 
is  proper  to  confine  our  investigations  to  small  numbers,  and 
from  observations  on  numbers,  we  deduce  the  following  prin- 
ciples for  finding  these  factors : 

1st.  That  any  number  ending  with  an  even  number,  or  a 
cipher,  can  be  divided  by  2. 

2d.  Any  number  ending  with  5  or  0,  is  divisible  by  5. 

3d.  If  the  right  hand  place  of  any  number  be  0,  the 
Avhole  is  divisible  by  10  ;  if  there  be  two  ciphers,  it  is  divisi- 
ble by  100;  if  three  ciphers,  by  1000,  and  so  on,  which  is 
only  cutting  off  those  ciphers. 

4th.  If  the  two  right  hand  figures  of  any  number  be  divis- 
ible by  4,  the  whole  is  divisible  by  4 ;  and  if  the  three  right 
hand  figures  be  divisible  by  8,  the  whole  is  divisible  by  8, 
and  so  on. 

5th.  If  the  sum  of  the  digits  in  any  number  be  divisible  by 
3  or  by  9,  the  whole  is  divisible  by  3  or  by  9. 

6th.  If  the  right  hand  digit  be  even,  and  the  sum  of  all 
the  digits  be  divisible  by  6,  then  the  whole  is  divisible  by  6. 

7th.  A  number  is  divisible  by  11,  when  the  sum  of  the  1st, 
3d,  5th,  &c.,  or  all  the  odd  places,  is  equal  to  the  sum  of  the 
2d,  4th,  6th,  &c.,  or  of  all  the  even  places  of  digits. 

8th.  If  a  number  cannot  be  divided  by  some  quantity  less 
than  the  square  root  of  the  same,  that  number  is  a  prime, 
or  cannot  be  divided  by  any  number  whatever. 

9th.  All  prime  numbers,  except  2  and  5,  have  either  1,  3, 


58  ELEMENTARY  ALGEBRA. 

7,  or  9,  in  the  place  of  units;  and  all  other  numbers  are 
composite  numbers,  and  can  be  divided. 

(Art.  28.)  The  multiple  of  a  number  is  some  exact  number 
of  times  that  number.  Thus,  6  is  a  multiple  of  2,  3 ;  12  is 
also  a  multiple  of  2,  3 ;  but  not  so  small  a  multiple  as  6  is, 
therefore  6  is  the  least  common  multiple  of  2  and  3. 

The  least  common  multiple  of  several  numbers  is  the  least 
number  that  is  divisible  by  these  numbers  without  a  remainder, 

A  COMMON  MULTIPLE  IS  FOUND  BY  MEANS  OF  PllIME   FACTORS. 

For  example,  find  the  least  common  multiple  of  the  num- 
bers 24,  20,  and  15. 

That  is,  find  the  least  number  which  is  divisible  by  24,  20, 
and  15.  First  find  the  prime  factors  to  these  numbers,  (Art. 
27),  (2,  2,  2,  3,)  (2,  2,  5,)  (3,  5). 

That  the  number  required  may  be  divisible  by  the  first 
number  24,  it  must  have  all  the  factors  in  that  number;  that 
is,  2,  2,  2,  3 ;  and  to  be  divisible  by  the  second  number,  20, 
it  must  contain  the  factor  5 ;  putting  in  this  factor  we  have 
2,  2,  2,  3,  5.  This  number  is  divisible  also  by  1 5,  because  it 
contains  the  factors  3,  5.  The  least  common  multiple  re- 
quired is,  therefore,  120. 

The  least  common  multiple  of  the  numbers  3,  7,  19,  is 
their  product,  because  the  numbers  are  prime,  and  there  is 
no  common  factor  that  can  be  cast  out. 

On  these  principles  the  following  rule  for  finding  the  com- 
mon multiple  will  be  easily  comprehended : 

Rule. —  Write  the  numbers  one  after  the  other,  and  draw  a 
line  beneath  them ;  then,  take  any  prime  number  which  will  di- 
vide two  or  more  of  them  vnthout  remainder,  and  divide  all  the 
numbers  that  will  so  divide — writing  the  quotients  beneath,  and  cdl 
the  numbers  that  are  not  divisible  by  it.  Find  a  prime  numher 
that  will  divide  two  or  more  numbers  in  this  second  line,  and 
proceed  as  before.     Continue  the  operation  until  there  are  no  two 


FACTORING. 


59 


numbers  left  having  a  common  divisor :  then,  multiply  all  the 
divisors  and  remaining  numbers  together,  and  their  product  will 
he  the  least  common  multiple  sought. 


EXAMPLES. 

1 .  Let  it  be  required  to  find  tlie  least  common  multiple  of 
12,  15,  7,  18,  3,  5,  and  35. 


7 

12, 

15,     ' 

r,    18, 

3, 

5, 

35, 

6 

12, 

15,      ] 

i,      18, 

3, 

5, 

5, 

3 

12, 

3,      1 

I,     18, 

3, 

1, 

1, 

2 

4, 

1, 

I,       6, 

1, 

1, 

1, 

2, 

1, 

I,       3, 

1, 

1, 

1, 

7X5X3X3X2X2=1260. 

2.  Find  the  least  number  that  can  be  divided  by  9,  1 2,  1 6, 
24,  36,  without  remainders.  Ans.  144. 

3.  Find  the  least  number  that  is  divisible  by  each  of  the 
nine  digits.  Ans.  2520. 

4.  Find  the  least  number  divisible  by  75,  50,  15,  20,  30, 
and  45.  Ans.  900. 

(Art.  29.)  A.  prime  quantitij  in  Algebra,  like  z,  prime  num- 
ber, is  divisible  only  by  itself  and  unity.  Thus,  a,  h,  a-\-b, 
are  prime  quantities  ;  and  ah,  and  ab-\-ac,  are  composite  quan- 
tities, the  first  is  composed  of  the  factors  a,  h,  and  the  other  of 
the  factors  a,  and  {b-\-c). 

The  prime  factors  of  a  purely  algebraic  quantity  consisting 
of  a  single  term,  are  visible  to  the  eye,  and  this  is  one  of  the 
principal  advantages  of  an  algebraic  expression. 

Thus,  in  the  expression  obex,  we  perceive  at  once  the  prime 
factors,  a  b,  c,  and  x;  the  expression  o?b'^x  has  three  prime 
factors,  each  equal  to  a,  two  prime  factors  equal  to  b,  and  one 
equal  to  x. 


60  ELEMENTARY  ALGEBRa. 

(Art.  30.)  When  the  algebraic  expression  is  a.  polynomial y 
and  has  prime  factors  that  are  monomials,  such  monomial 
factors  are  visible,  as  in  the  following  expressions : 

Factors. 

1.  x-]rax  (^\-\-a)x 

2.  am-\-an-]r(ix  {m-\-n-\'x)a 

3.  bc^-^-hcx-^rhcy  (^c-\-x-\-y)hc 

4.  4a;2+6a:y  (2a;+3y)2a: 

Thus  in  the  first  expression,  x  is  visible  in  every  term,  it 
is,  therefore,  a  common  factor  to  every  term,  and  ( 1  -\-a)  is 
the  other  factor,  and  the  product  of  these  two  factors  makes 
the  expression  ;  and  so  for  the  other  expressions.     • 

The  examples  in  division  (Art.  23),  are  analagous  to  these, 
except  that  in  that  article  the  divisor  is  given,  and  may  not 
be  contained  in  every  term,  as  in  example  7,  (Art.  23). 

(Art.  31.)  When  all  the  prime  factors  composing  any 
algebraic  expression  consist  of  binomials  or  polynomials,  they 
are  not  visible  in  the  expression  like  a  monomial,  and  we  can 
find  them  only  from  our  general  knowledge  of  algebraic 
expressions. 

For  instance,  the  prime  factors  in  the  expression  (a^-|-2ad 
'\-l?)  we  know  to  be  (a+5)  and  {a-\-h)  by  (Art.  16),  and 
all  other  expressions  that  correspond  to  a  binomial  squared, 
is  immediately  recognized  after  a  little  experience  in  algebraic 
operations. 

Also,  any  expression  which  is  the  difference  of  two  squares, 
as  (a^ — 52^  jg  instantly  recognized  as  the  product  of  the  two 
prime  factors,  («-!-&),  and  (a — h),  (Art.  17). 

The  expression  ax-^ay-\-hx-\-hy  can  be  resolved  into  two 
prime  factors,  by  inspection,  thus,  a{x-\-y)-\-b{x-{-y)  is 
merely  a  change  in  the  form  of  the  expression.  Now  put 
{x-\-y)  —  S.  Then  the  next  change  is  aS-]rbS;  the  next  is 
{a-\-b)S,  Restoring  the  value  of  S,  we  have  (a-\-b){x'{-y) 
for  the  prime  factors  i:^  the  original  expression. 


FACTORING.  61 

(Art.  32.)  Any  trinomial  expression  in  the  form  of  ax^-{' 
hx-\-c,  can  be  resolved  into  two  binomial  factors  ;  but  the  art 
of  finding  tlie  factors  is  neither  more  nor  less  than  resolving 
an  equation  of  the  second  degree,  a  subject  of  great  impor- 
tance and  some  difficulty,  which  will  be  examined  very  closely 
in  a  subsequent  part  of  this  work ;  therefore  it  is  improper  to 
treat  upon  this  subject  at  present.  (See  Art.  95). 

(Art.  33.)  Common  multiple,  and  least  common  multiple, 
have  the  same  signification  in  Algebra  as  in  Arithmetic,  and 
are  found  by  the  same  rule,  except  changing  the  words  num- 
ber and  numbers  in  the  rule  for  quantity  and  quantities. 

Or,  we  may  take  the  following  rule  to  find  the  least  common 
multiple  in  algebraic  quantities. 

Rule  . — 1 .  Resolve  the  numbers  into  their  prime  factors. 

2.  Select  all  the  different  factors  which  occur,  observing,  when 
the  same  factor  has  different  pmvers,  to  take  the  highest  power. 

3.  Multiply  together  the  factors  thus  selected,  and  their  pro- 
duct will  be  the  least  common  multiple. 

EXAMPLES. 

1 .  Find  the  least  common  multiple  of  Wx^y,  and  1 9.a^I^x. 
Resolving  them  into  their  prime  factors, 

8aVy=2'^Xa2Xa;2Xy 
nan'x=^^''Xa^XxXb^X^ 
The  different  factors  are  ^^,  d^,  x^,  y,  b^,  3,  and  their  product 
is  ^\c?ly^x'^y,  which  is  the  least  common  multiple  required. 

2.  Required  the  least  common  multiple  of  27a,  155,  9a5, 
and  3a2.  Ans.  I'iba'^b. 

3.  Find  the  least  common  multiple  of  (a^ — x^),  4(a — x)^ 
(a+ar).  Ans.  M^a^ — x^), 

4.  Required  the  least  common  multiple  of  a?{a—^x),  and 
ax\Gi? — 7?),  Ans.  a^x^a? — x*). 


62  ELEMENTARY  ALGEBRA. 

5.  Required  the  least  common  multiple  of  x^(x — y),  aV, 
and  12axf.  "  Am.  12aV/(a;_y). 

6.  Required  the  least  common  multiple  of  10aV(a — 5), 
15a^(a-\-b),  and  nioJ^—l^).  Ans.  60a^a^(a^—P), 


ALGEBRAIC    FEACTIONS. 

The  nature  of  fractions  is  the  same,  whether  in  Arithmetic 
or  Algebra,  and  of  course  those  who  understand  fractions  in 
Arithmetic,  can  have  no  difficulty  with  the  same  subject  in 
Algebra. 

(Art.  33.)  A  fraction  is  one  quantity  divided  by  another 
when  the  division  is  indicated  and  not  actually  performed. 

Hence  every  fraction  consists  of  two  parts,  the  dividend  and 
divisor,  which  take  the  name  of  numerator  and  denominator. 

The  numerator  is  written  above  a  Une,  and  the  denominator 

below  it,  thus,  _,  and  is  read,  a  divided  by  h. 
h 

For  illustration,  we  may  consider  any  simple  fraction  as  | ; 

here  we  consider  one  or  unity  divided  into  5  parts,  and  3  of 

these  parts  are  taken.     The  5  denotes  the  parts  that  the  unit 

is  divided  into,  hence  it  is  properly  named  denominator,  and 

the  3,  numbers  the  parts  taken,  and  is,  therefore,  properly 

called  the  numerator.     So  in  the  fi-action  -,  b  denotes  the 

b 

parts  into  which  unity  is  divided,  and  a  shows  the  number  of 

parts  taken. 

In  a  numeral  fraction,  as  |,  it  is  evident  that  if  we  double 

both  numerator  and  denominator,  we  do  not  chanofe  the  value 

of  the  fraetien ;  thus>  f  is  the  same  part  of  the  whole  unit  as 


ALGEBRAIC  FRACTIONS.  63 

|,  and  thus  it  would  be  if  we  multiplied  by  any  other  num- 
ber ;  and  conversely,  we  may  divide  both  numerator  and  de- 
nominator by  the  same  number,  without  changing  tha  value 
of  the  fraction.  Hence,  if  any  fraction  contains  any  factors 
common  to  both  numerator  and  denominator,  we  may  sup- 
press them  by  division,  and  thus  reduce  the  terms  of  the  frac- 
tions to  smaller  quantities. 

Hence,  to  reduce  fractions  to  lower  terms  when  possible, 
we  have  the  following  rule : 

Rule  . — Divide  both  terms  by  their  greatest  common  divisor. 
Or,  resolve  the  numerator  and  denominator  into  their  prime  fac- 
tors, and  then  cancel  those  factors  common  to  both  terms. 

EXAMPLES. 

1.  Reduce to  its  lowest  terms. 

21a6c-2 

Here  labc  is  the  common  divisor,  and  dividing  according 

2a5^ 

to  the  rule,  gives  ,  the  fraction  reduced. 

3c 

2.  Reduce  — ^_ Ans.     ~. 

55a^x^  5x 

3.  Reduce to  its  lowest  terms Ans. 

18a5  33 

4.  Reduce  H^^^  to  its  lowest  terms.  .     .     .     Ans.  ^l. 

2\ax^  3 

6.  Reduce  il^^  to  its  lowest  terms.      .       Ans.  ?^. 

^    rf  A        Bla^b—Q^a^b"^  ,    .,    , 

6.  Reduce  — --— to  its  lowest  terms. 

2>Qa'b^—2ab 

Ans. 

12a'b-~3 

7.  Reduce  i^!z:i^'  to  its  lowest  terms.        Ans.  K^ZI^. 

3(a4-a:)  3 

8.  Reduce     ^     ^  ■  to  its  lowest  terms.       .     Ans.  _— . 


64  ELEMENTARY   ALGEBRA. 

^ \ 

9.  Eeduce to  its  lowest  terms.     .     .     Ans. 


X- 


10.  Reduce \ to  its  lowest  terms.        Ans.  — ~     . 

acx-f-abx  ac-rob 

11.  Divide  icV'+ar'y' by  or^y-fory^.         .     .     .    Ans,  ^. 

a 

12.  Divide4a+45by2a*— 2S* Ans.—. 

a — b 
« 

13.  Divide  %^ — 2n^  hj  n^ — 4w-{"4.        .     .     .    Ans. 


(Art.  34.)  Fractions  in  Algebra,  as  in  Arithmetic,  may  be 
simple  or  complex,  proper  or  improper,  and  the  same  defini- 
tions to  these  terms  should  be  given,  as  well  as  the  same  rules 
of  operation ;  for  in  fact  this  part  of  Algebra  is  but  a  gener- 
alization of  Arithmetic,  and  in  some  cases  we  give  arithmeti- 
cal and  algebraical  examples  side  by  side. 

A  mixed  quantity  in  Algebra  is  an  integer  quantity  and  a 
fraction ;  and  to  reduce  these  to  improper  fractions,  we  have 
the  following  rule : 

Rule  . — Multiply  the  integer  hy  the  denominator  of  ihefrac- 
tioUj  and  to  the  'product  add  the  numerator,  or  connect  it  with  its 
proper  sign  -\-  or  — ;  then  the  denominator  being  set  under 
this  suniy  will  give  tJie  improper  fraction  required* 

EXAMPLES. 

1.  Reduce  2f  and  a-\--  to  improper  fractions. 

Ans.  V  and  ^±1. 

These  two  operations,  and  the  principle  that  governs  them, 
are  exactly  ahke. 

2.  Reduce  6}  and  a-f—  to  improper  fractions. 

A7i>.  Y  and  5i+:f!. 


ALGEBRAIC  FRACTIONS.  65 

3.  Reduce  7|  and  ax-\--  to  improper  fractions. 

c 

4.  Reduce  3 — |  and  x^ — -  to  improper  fractions. 

Ans.  4  and  ^tZ^. 

y 

6.  Reduce  y — \-\-ZIIK  to  a  fractional  form.      Ans.  ^     ^. 
1+y  y+1 

6.  Reduce  x-\-y-\- to  tlie  form  of  a  fraction. 

x-{-y 

Ans.  ^+^^y+y^+g 

h  '  ^+^ 

7.  Reduce  4+2a;-j--  to  an  improper  fraction. 

c 

8.  Reduce  5x — — Z_  to  an  improper  fraction. 

3^2 2Q 

9.  Reduce  3a — 9 — to    a    simple    fraction. 

a-1-3  ^ 


-4n5. 


(a-l-37 

The  converse  of  this  operation  must  be  true,  and,  there- 
fore, to  reduce  an  improper  fraction  to  a  mixed  quantity,  we 
have  the  following 

Rule  . — Divide  the  numerator  by  the  denominator,  as  far  as 
possible,  and  set  the  remainder,  (if  any),  over  the  denominator 
for  the  fractional  part ;  the  two  joined  together  with  their  proper 
sign,  will  be  the  mixed  quantity  sought. 

EXAMPLES. 

1 .  Reduce  y  and  to  mixed  quantities. 

X 

Ans.  5 1  and  a-\-i* 

2.  Reduce  \^  and  —L —  to  mixed  quantities. 

bx 
An^.  2|  and  a-\ — • 


66  ELEMENTARY   ALGEBRA. 

3.  Reduce  — ^        "*      to  mixed  quantities. 

y 

Ans.  5a-\-^        « 

2^2 QJ2 

4.  Reduce 1-  to  a  whole  or  mixed  quantity. 

a — b 

Ans.  2a-{-2b. 

5.  Reduce lil —  to  a  mixed  quantity.      Ans.  3a-\-—. 

bdf-  bet 

6.  Reduce  ^  ~^^    '      to  a  mixed  quantity. 

a 

Ans.  a-\-h-\--~. 
a 

7.  Reduce X to  a  mixed  quantity, 

4a 

Ans.  Sa-\-l——. 
4a 

(Art.  35.)  A  fraction  is  an  expression  for  unperformed 
division.  Thus,  2  divided" by  5,  is  written  |.  The  double  of 
this  is  f ,  3  times  f  is  |,  &c.  That  is,  to  multiply  a  fraction 
by  any  number,  we  multiply  tlie  nwnerator  of  the  fraction  hy 
the  nwmhery  vnthovt  changing  the  denominator. 

The  nature  of  division  is  the  same,  whatever  numbers  rep- 
resent the  dividend  and  divisor.  Hence,  for  the  sake  of 
simplicity,  let  us  consider  the  result  of  dividing  24  by  6. 
Here  24  is  the  dividend  and  6  the  divisor,  and  the  division 
expressed  and  unperformed,  must  be  written  \* ,  and  the  value 
of  this  expression,  or  quotient,  is  4.  Now  observe,  that  we 
can  double  the  quotient  by  doubling  24,  or  by  taking  the  half 
of  6.  We  can  find  3  times  the  value  of  this  quotient,  by 
multiplying  the  numerator  24  by  3,  or  hy  dividing  the  denom- 
inator 6  by  3. 

Hence,  to  multiply  a  fraction  by  a  whole  number,  we  have 
the  following  rule : 


ALGEBRAIC  FRACTIONS. 


at 


Rule  . — Multiply  the  nvmeraior  hy  the  whole  number ;  or, 
when  you  can,  divide  the  denominator  by  the  whole  nianber. 


EXA 

1.  Multiply  f  by  5. 

2.  Multiply  I  by  3.,      . 

3.  Multiply  If  by  4.     . 

4.  Multiply  j\  by  100. 

5.  Multiply  ij  by  18. 

6.  Multiply  i-f  by  19. 

7.  Multiply  1  by  24.     . 

8.  Multiply  ii  by  105. 

9.  Multiply  f  by  63.    . 

10.  Multiply -^  by  c. 
b 

(Art.  36.)  When  we  multiply  a  fraction  by  its  denomina- 
tor, we  merely  suppress  the  denominator.  Thus,  multiply  ^ 
by  3,  the  result  is  1,  the  numerator  of  the  fraction  ;  multiply 
I  by  5,  and  we  have  2,  the  numerator  for  the  product. 


MPLES. 

.     Ans.   y=2| 

.     Ans.     -1=1^ 

Ans.     |f=3ii. 

.      Ans.  V4°=64f 

,     .     .      Ans.     7-J 

.     .     .      Ans.  5if 

.    \      Ans.     56 

.     .      Ans.     85 

.     .      Ans.     27 

.     .      Ans.      — . 

EXAMPLES. 

1.  Multiply -f  by  7 Ans.  3. 

2.  Multiply  -hyb Ans.  a. 

b 

3.  Multiply  yV  by  1 1 ■^^'  4. 

4.  Multiply -^  by  11 Ans.  x. 

5.  Multiply  Jr  by  55 Ans.  Sax. 

5b 

St- 

6.  Multiply  _  by  7 Ans.     3x. 


7.  Multiply  ?^^  by  20. 


Ans.  6a — 2x. 


68  ELEMENTARY  ALGEBRA. 

8.  Multiply  ^l^^H^by  6bx,       .     .     .       Ans.  42ax—^b. 

Sbx 

9.  Multiply  t-\--  by  6 Ans.  3x+2x. 

A)  O 

10.  Multiply  ?f +?f  by  3 Am,  2ar+^. 

o        ^  2, 

U.  Multiply  3|  by  3 Ans.  10. 

(Art.  37.)  As  a  fraction  is  an  expression  for  unperformed 
division,  we  may  express  the  division  of  %\  by  6|-,  in  tlie 
following  form : 

But  this  is  certainly  a  complex  frcLction ;  so  are  -1  and 

complex  fractions ;  hence  complex  fractions  may  be  defined 
thus : 

A  complex  fraction  is  (yne  in  which  the  numerator  or  denom.' 
inator,  or  loth,  are  fractions  or  mixed  quantities. 

To  simplify  a  ccmplex  fraction,  we  multiply  loth  numerator 
and  denominator  hy  the  denominators  of  the  fractional  parts  : 
or  by  their  product,  or  by  their  least  common  multiple. 

For  example,  let  us  simplify  the  fraction  -?.  If  we  mul- 
tiply  both  numerator  and  denominator  by  2,  the  numerator 
will  contain  no  fraction,  and  the  result  will  be Multiply 

3 

numerator  and  denominator  of  this  fraction  by  3,  and  the 
denominator  will  contain  no  fraction ;  and  the  final  result  will 

be  _»  a  simple  fraction,  equal  in  value  to  the  complex  fraction. 

But  we  could  have  arrived  at  this  result  at  once,  by  multi- 
plying both  terms  by  6,  the  product  of  2»3.  Hence,  the  rule 
just  given. 


ALGEBRAIC  FRACTIONS.  69 

EXAMPLES. 

1.  Reduce  -J  to  a  simple  fraction Ans»  |f . 

5 

2.  Reduce  —  to  a  simple  fraction.     .     .     .    Ans.  |=1|. 

m 

3.  Reduce to  a  simple  fraction.      .    Ans,  . III — . 

,     c  nbd — en 

4.  Reduce to  a  simple  fraction.      .     .     .     Ans.  -xt* 

h  36 

5.  Reduce  ifJif  to  a  simple  fraction.     .    .  Ans.  J^_f_. 

y  4a;-f2y 

a 

6.  Divide  -  by  -,  that  is,  simplify  the  complex  fraction  —' 

b       d  _ 

d 

Here  the  division  is  expressed,  but  unperformed,  and  by 

the  rule  to  simplify  the  fraction,  we  find  its  value  to  be  — . 

be 

From  this  result  -we  can  draw  a  rule  for  dividing  one  frac- 
tion by  another ;  and  the  rule  here  indicated,  when  expressed 
m  words,  is  the  rule  commonly  found  in  Arithmetic. 

7.  Simplify  the  fraction Ans. 


,  1  ac-fl 

8.  Simplify  the  fraction Ans. 


m  n-{-m 


rm  ELEMENTARY  ALGEBRA. 


MULTIPLICATION   OP  FRACTIONS. 

(Art.  38.)  We  have  already  given  a  rule  to  multiply  a 
fraction  by  a  whole  number;  (Art.  35) ;  but  when  two  frac- 
tions are  multiplied  together,  the  result  is  the  same,  whichever 
we  consider  the  multiplier.     That  is,  |  multiplied  by  4,  and 

4  multiplied  by  |,  is  the  same  product.     Also,  -  multiplied 

b 

by  X  is  _— ,  therefore,  x  multiplied  by  -  is  also  —.    Hence  to 
b  b  b 

multiply  a  quantity  by  a  fraction,  observe  the  following  rule  : 

B  u  L  E . — Multiply  the  quantity  by  the  numerator  of  the  frac- 
iian,  and  set  the  denominator  under  the  result. 


EXAMPLES. 

1.  Multiply  7  by  f. 

.     .     .     Ans.  V- 

2.  Multiply  a  by  -. 

.     .     .     Ans.^J.. 

y 

3.  Multiply  5  by  4. 

.     .     .     Ans.  Y. 

Now,  in  this  last  example  write  (me  under  the  5,  which  will 
give  it  a  fractional  form  without  changing  its  value.  Then  it 
will  be  f-  X  4  ;  and  if  we  multiply  the  numerators  together,  and 
the  denominators  together,  we  have  Y»  as  before.  Again, 
we  may  take  \  and  multiply  both  numerator  and  denom- 
inator by  any  number,  say  3,  and  we  have  y,  which  is 
really  5  as  at  first.  We  have  now  to  multiply  y  by  4,  and 
if  we  multiply  numerators  and  denominators  as  before,  we 
shall  have  |^  for  the  product,  which  is  in  value  Y»  as  it 
ought  to  be. 

4.  Multiply  ahyi.    . Prod.  ^. 

d  d 


MULTIPLICATION   OF  FRACTIONS.  TL 

5.  Multiply  ^  by  1 Frod.  1'. 

la  d 

As  a=^  we  can  thus  change  the  form  of  the  first  factor 
n 

without  changing  its  value,  then  the  example  will  be  to 

Multiply!??  by  £.  .% Prod.'"^^^!, 

n        d  nd      d 

From  these  examples  we  have  the  following  rule  for  multi- 
plying fractions  together : 

E,  u  L  E. — Multijply  the  numerators  together  for  a  new  nume- 
rator ^  and  the  denominators,  for  a  new  denominator. 

N.  B.  Equal  factors  in  numerators  and  denominators  may 
be  canceled  out,  which  will  save  the  reduction  of  the  product 
to  lower  terms. 

To  find  such  equal  factors,  separate  the  quantities  into  their 
prime  factors  (Art.  27),  before  multipHcation. 

EXAMPLES. 

1.  Multiply  ??  by  ^ Ans.  ii^. 

2.  Multiply  2?  by  ?-^ Ans.     -?. 

^  ^  by    ^  dx  5x 

3.  Multiply  ?^  by  ^ Ans.      1 

^  "^  lOy    "^  9a;  6 

A    -KIT  n-  1    « — ^  T,     25a; — 25  ,1  .5 

4.  Multiply by  — — _  by .     .     Ans. 


5       •'     a'—b^      "  x—\  a-\-h 

In  this  example  we  separate  the  second  fraction  into  its 
prime  factors,  (Art.  27),  and  the  operation  stands  thus  : 
a—h^     25(a;— 1)      ^     1 
5        {a-\-b){a—b)     x—\ 

Suppressing  all  the  factors  which  are  found  common  in  the 

5 

numerator  and  denominator,  and  the  result  is ,  ans. 

a+b 


T9 


5.  Multiply 


ELEMENTARY  ALGEBRA. 

X 


a-^x 


,  and  — ^   together.     Ans.  - 

ar  a — x 


6.  Multiply  ??  by  2^.     .     . 

7.  Multiply  1^^  by  ?^.  .     . 

8.  Multiply  ?f  by  ^X^'. 


Ans,     ^, 
xz 


Ans. 

Atis. 


4a 
z 

a 


9.  Multiply  — -  into  — ^.    .     . 
X  z 

10.  Multiply  -,  —,  ^  together. 

11.  Multiply  (^±^  by     ;^^      . 

^•^30         ^    3(a+x) 

12.  Multiply  ?^±i^  by  ??.       . 

13.  What  is  the  product  of  -—,  l^  and  ??      Ans.  12:j 

y       2a  X 

14.  What  is  the  product  of  — _  into  ^f^lt 

^  2>b-\-c  5ab 


Ans. 


X 

9ax 


2b 
Ans.    a^x 

Ans.  9ax. 


18 
Ans.  ^±5^ 


Ans. 


4ac—2bc 


I5b'-i-5bc 
15.  Multiply  5-I-?5  by  1      .....     Ans.  ^^^fl. 


^ 7,2 

16.  Multiply  fZZ-  by 
be 


X 


Ans. 


bh-^b^ 


17.  Multiply  ^!i:^  by  ^ Am.  i^Z^, 

2y  a-\-x  ■  y 

18.  Multiply  —t,  JL.  and  -^ Ans.  a. 

X       x-\-y  X — y 


MULTIPLICATION  OF  FRACTIONS. 


19.  Multiply  3a,  fill,  and  ^  together. 
2a  a-i-b 


20.  Multiply  ^"^""^^  by  —2^. 

^"^        14         ''   2a^—3x 

21.  Multiply  -Jf-   by  1^^=??. 

^•^    6;^;— 10     •"         2a; 


Ans. 


Am. 


73 


3(a:^~-l) 

2(a+6)  • 

3ax — 5a 

4a;2~6" 

Ans.  £f. 


22.  Multiply  ^-  by  i- Am,!, 


DIVISION    IN    FRACTIONS. 

(Art.  39.)  When  we  multiply  a  fraction  by  a  whole  num- 
ber, we  multiply  the  numerator  by  that  number,  or  divide 
the  denominator,  (Art.  35) ;  and  as  division  is  the  converse 
of  multiplication,  therefore,  conversely,  when  we  divide  a 
fraction  by  a  whole  number,  we  divide  the  numerator  (when 
possible),  or  multiply  the  denominator  by  that  number. 
Thus,  -f  divided  by  3,  would  be  |,  and  divided  by  4,  would 
be  -y^ ;  jn  the  first  case  the  division  is  actually  performed,  in 
the  second  it  is  only  expressed. 


EXAMPLEJ 

3. 

1.  Divide    |  by  3 

Ans.    J. 

2.  Divide    f  by  9 

^^.  ^. 

3.  Divide    f  by  6 

.....     Ans,     \. 

4.  Divide  j-f  by  13 

^^'^' 

a.  Divide  tV  by  8 

^«*.  iV 

74  ELEMENTARY  ALGEBRA. 

6.  Divide  —  by  3c Ans.  . — 

b     ^  ch 

In  this  example  we  divide  first  by  3,  and  that  quotient  by  c. 

7.  Divide  ?^by^ Ans.  ^^. 

(Art.  40.)  Let  us  now  consider  division  when  the  divisor 
is  a  fraction.  We  must  now  go  back  to  the  elementary  prin- 
ciple of  division.  It  is  the  art  of  discovering  how  many  times 
a  number  or  qtuintity,  called  the  divisor,  can  be  subtracted  from 
another  number  or  quantity  of  the  same  kind,  called  the  dividend. 

For  example,  we  require  the  division  of  6  by  -J.  The  un- 
disciplined and  inconsiderate  often  understand  this  as  demand- 
ing the  third  of  6  ;  but  it  is  not  so,  it  is  demanding  how  many 
times  -J  is  contained  in  6,  or  how  many  times  -J  can  be  svh- 
traded  from  6. 

To. arrive  at  the  true  result,  we  consider  that  -J  is  contained 
in  1  three  times  ;  therefore,  it  must  be  contained  in  6,  18  times. 

Now,  |-  must  be  contained  in  6,  9  times ;  and  we  may  arrive 

6*3 

at  this  result,  thus, 

2 

Again,  suppose  we  divide  the  number  a  by  |. 

The  divisor  |  is  contained  in  one  unit  7  times,  therefore,  it 
is  contamed  in  a  units,  la  times. 

To  make  this  more  general,  we  will  suppose  the  denomina- 
tor of  the  divisor  to  be  any  other  number  as  well  as  7  ;  there- 
fore, suppose  it  n,  the  quotient  will  then  be  na.  To  make 
the  example  still  more  general^  let  us  suppose  a  to  be  divided 

by  - ,  m  being  a  wliol^  njimber. 
n 

The  divisor  can  be  resolved  into  two  factors,  -  and  m.     Di- 

n 

viding  a  by  the  factor  -,  we  have  already  shown  the  quotient 


DIVISION  IN  FRACTIONS.  75 

to  be  na ;  dividing  this  by  tbe  whole  number  m,  (Art.  39), 

the  residt  must  be  — -. 
m 

This  shows  that  when  the  divisor  is  a  fraction,  the  quotient 

is  found  hj  the  following  rule  : 

Rule  . — Multiply  the  dividend  (whatever  it  may  he)  hy  the 
denominator  of  the  divisor^  and  divide  thai  product  hy  the 
numerator. 

In  the  result  last  given,  let  the  dividend  a  be  a  fraction 

_,  aud  in  the  place  of  a  write  _. 
d\  ^  d 

Then  the  problem  will  be  to  divide  £.  by  —,  that  is,  one  frac- 

d        n 

tion  by  another,  and  the  result  must  be 

c 
n- 

d 


This  is  a  complex  fraction,  and  simplifying  it  (by  Art.  37), 

1         nc 
we  have  — 
md 

From  this  result  we  draw  the  following  rule  for  dividing 
one  fraction  by  another  : 

Rule  . — Invert  the  terms  of  the  divisor,  and  proceed  as  in 


(Art.  41.)  For  the  purpose  of  illustrating  the  nature  of 
an  equation,  and  showing  the  power  and  simplicity  of  alge- 
braic operations,  we  will  arrive  at  this  rule  by  another  course 
of  reasoning. 

Let  us  again  consider  the  nature  of  division,  and  for  this 
purpose,  divide  32  by  8. 

Divisor.    Dividend.    Quotienl. 

8     )     32     (     4 


76  ELEMENTARY  ALGEBRA. 

Here  it  is  visible  that  the  product  of  the  divisor  and  quo- 
tient is  equal  to  the  dividend ;  and  this  is  a  general  principle, 
true  in  every  possible  case. 

Now  let  us  divide  -  by  -.     There  will  be  a  certain  quotient 
b       d 

which  we  will  represent  by  Q.     Then  the  product  of  the 

divisor  and  quotient  will  be  equal  to  the  dividend ;  that  is,  we 

shall  have  the  following  equation  : 

cQ     a 

Coth  members  of  this  equation  are  fractional,  and  if  we 
multiply  the  first  member  by  d,  the  denominator,  the  product 
will  be  the  numerator,  (Art.  35)  ;  but  if  we  take  d  times  one 
member,  we  must  take  d  times  the  other  member,  to  preserve 
equality.     (Ax.  3). 

Therefore,  multiplying  by  d,  we  have 

e*  ad 

Dividing  both  members  by  c,  then  Q  will  stand  alone. 

And «=^ 

oc 

This  equation  shows  that  when  we  divide  one  fraction  by 
another,  the  value  of  the  quotient  is  found  by  inverting  the 
terms  of  the  divisor,  and  then  multiplying  the  numerators 
together  for  a  new  numerator,  and  the  denominators  together 
for  a  new  denominator  ;  or  more  briefly,  we  say 

Invert  the  terms  of  the  divisor,  and  proceed  as  in  multiplication. 

EXAMPLES. 

1.  Divide  f  by  f Ans.  \l=^\^. 


DIVISION  IN  FRACTIONS. 


77 


2.  Divide  by  _. 

1— a     "^    5 

3.  Divide  ?f  by  ^. 

a5  a6 


4.  Divide  .  by  — 

ah  ah 

5.  Divide  'l^^-t  byf±f!f. 

5a5/  5as^ 

6.  Divide  ^  by  _£_.       . 

c  a-\-h 

7.  Divide  ~  by  -. 

a  c 

8.  Divide by  .. 

a — X         (t — ^ 


Ans. 


.     Ans.  — 

3y 

3a— 5 


.    Ans. 


Ans. 


Ans. 


2d' 
Trxj-a" 
4b-\-ax 
(a+hf 


Operation, 

15ah^(a-\.x)(a~^x) 
a — X            lOac 

>...    3J(«+.) 

2c 

9.  Divide 

a^ — x^          a — X 

.       Ans.       2«+-     . 

10.  Divide 

14^-3  .      lO.r-4 
6          -^        25 

102'— 4 

11.  Divide 

^^-^^by^.       .     . 

.        9^—3 
•     1     t     Ans. 

5        ^5         •     ' 

x 

12.  Divide 

60.-7,      o:-! 

A..    1S.;-21 

x+l      ^3 

x^—\ 

13.  Divide 

16a^  by  ^"^ 
5        -^    15       '     *     * 

.     .     .     .   Ans.  12a. 

14.  Divide 

6,+4         3.-I-2    ^ 

,     ,     .     ,     Ans    ^y. 

5         -^       4y 

5 

15.  Divide 

7-^  by  ^^ 

..        21 

3      ^    6 

6a; 

*  Separate  into  factors  wherever  separation 

is  obvious. 

78 


ELEMENTARY  ALGEBRA. 


.    16.  Divide  «±1  by  2^.     .     . 
6        -^    3 

.    1 7.  Divide  by  - .       .     . 

x—l     ^  2 

18.  Divide  ^ti:^y±l  by  ^"Z^ 
ab  he 


19.  Divide 


im; 


by 


in-\-n 


5x 


20.  Divide  _  by  — .  . 

21.  Divide  ^  by  ^i?. 

Set?      -^    4d 


22.  Divide 


a:-*— 5^ 


a:"-^— 26a?-l-&2 


by 


x^-\-bx 
X — b 


Ans. 
Ans. 


4a 

2 

x—\ 


Ans.  f^=fZ 
a 

^?^5.  2m — 27i 

2a 
a; — 5» 


Atis.  x-\- 


Qcrx 
a: 


Operation,  (^!±?Mz?)  >< --^- - 


oNrf 

X 


ADDITION    OF    FRACTIONS. 

(Art.  42.)  When  fractions  have  a  common  denominator, 

they  can  be  readily  added  together  by  adding  their  numera- 

4  7 

tore,  because  f  and  ^  is  obviously  ^,  and  _  and  -  is  obviously 

n         n 

11        a      J  6  .    a-\-h    - 
. — ,  or  -  and  -  is  — \ — ,  &c. 
n        n         n        n 

But  when  the  denominators  are  unlike,  we  cannot  directly 
add  the  fractions  together,  because  we  cannot  add  unlike 
things,  as  dollars  and  cents,  or  units  and  tens,  &c. 


ADDITION  OF  FRACTIONS.  79 

In  all  such  cases  we  can  only  indicate  the  addition  by  signs, 
unless  we  first  reduce  the  quantities  to  like  denominations,  or 
(as  applied  to  fractions),  to  common  denominators. 

We  shall  investigate  a  rule  for  the  addition  of  fractions 
through  the  medium  of  equations. 

For  example,  we  require  the  s^  of  ^,  |,  and  |.  By  the 
summary  process  of  Algebra,  we  pronounce  the  sum  to  be  S. 
Then  we  have  the  following  equation  : 

^=i+t+-f       (1) 

The  first  member  of  this  equation  is  a  symbol  merely ; 
and  in  the  second  member  the  addition  is  only  indicated^  not 
performed ;  and  to  perform  it,  the  fractional  form  of  the  equa- 
tion must  be  changed  to  whole  numbers,  or  the  denoihiriators 
made  common. 

If  we  multiply  every  term  of  both  members  hy  2,  the  first 
fraction  will  he  removed ,  (Art.  35),  and  the  equation  will  stand 
thus : 

25=1+1+1  (2) 

If  we  multiply  every  term  hy  3,  the  second  fraction  vrill  he 
removed,  and  the  equation  will  stand  thus : 

65=3+4+ V  (3) 

In  the  same  manner  we  can  remove  the  third  fraction  by 
multiplying  by  5  ;  then  we  have 

305=15+20+18         (4) 

Now,  if  we  divide  every  term  of  equation  (4)  by  30,  we 
shall  have 

Here  we  have  the  sum  5  equal  to  fractions  having  a  com- 
mon denominator,  and  that  common  denominator  is  the  pro- 
duct of  the  denominators  of  the  given  fractions  2,  3,  and  5. 

In  equation  (4),  we  may  add  the  numbers  15,  20.  and  18 
directly,  making  53,  and  the  equation  will  be 
305=53 


^  ELEMENTARY  ALGEBRA. 

Dividing  by  30,  and   S=^  or  Iff. 

Also,  the  sum  of  the  fractions  in  equation  (5)  is  /S^=f  f . 

That  the  operation  may  be  more  distinct,  we  will  require 

the  sum  of  the  literal  fractions  -,    t,  and  ?.. 

b     d  h 

Assume  >S^  to  be  their  sSn  as  before. 


a    c 


9  (1) 


Then /S'=_j___l- 

Remove  the  fractions  first,  by  multiplying  by  6,  then  by  c?, 
then  by  A  y  or  by  mvltiplying  the  whole  at  once  by  the  product 
Idh. 

Multiplying  by  b,  gives  bS=a^--\'it  (2) 

d      h 

Again  by  d,  gives        dbS==adJrcb-\-2^       (3) 

Again  by  A,  gives      hdbS=adh-\-cbh-\-gbd    (4) 

Dividing  both  members  of  equation  (4)  by  hdb,  and  we 

f^_^adh     cbh      gbd  r_x 

Mb'^hdb'^hdb  ^   ^ 

But  these  fractions  in  the  second  member  of  equation  (5), 
have  a  common  denominator,  and,  therefore,  it  need  not  be  writ- 
ten under  every  numerator,  if  it  be  written  under  their  sum. 

Ti„3      .     .     .     ^^M+,^t£^  (6) 

Here,  then,  we  have  the  sum  of  the  fractions  in  one  quantity. 

By  inspecting  the  second  member  of  equation  (5),  and 
comparing  it  with  the  original  fractions  to  be  added,  we  per- 
ceive that  the  numerator  of  the  first  fraction,  a,  is  multiplied 
by  the  denominators  of  the  other  fractions  ;  and  the  numera- 
tor of  the  second  fraction,  c,  is  also  multiplied  by  the  denom- 
inators of  the  other  fractions :  and  the  same  is  true  of  tlio 
third  fraction,  and  so  on. 


ADDITION  OF  FRACTIONS.  81 

The  common  denominator  is  made  up — or  is  tlie  product — 
of  all  the  denominators. 

Hence,  we  derive  the  following  rule  for  reducing  fractions  to  a 
common  denominator : 

Rule  . — Multiply  each  numerator  into  all  the  denominators 
except  its  mun,  for  the  new  numerators;  and  all  the  denominators 
together,  for  a  common  denominator. 

And  to  add  fractions,  we  have  the  following  rule  : 

Rule  . — Reduce  the  fractions  to  a  common  denominator ; 
and  the  sum  of  the  numerators,  written  over  the  common  denom- 
inator, will  he  the  sum  of  the  fractions. 

EXAMPLES. 

1.  Add  —,  —  and  -  together. 

^      '^  ^  ^^^    63a:4-30ar-f35a;     128a? 


105  105 

2.  Add  «  and  "-±-^ Ans.  ''J±^±l 


be 

X 

3.  Add  *,  -  and  t  together Ans.  .^•4-T?. 

2     3  4  ^2 

4.  Add  ^—  and  —  together Ans,  l?^tlli 

3  7       *  21 

5.  Add  and  together.   .     .     .      Ans.  ~- 

a-j-6  a — b  a~ — b"^ 

6.  Add  and  -J—,  together.    .     .     .      Ans.  — ^-^~ 

x-\-y  x—y  x'^—y'- 

3x    ^b 

7.  Reduce  — ,  — ,  and  d,  to  fractions  having  a  common 

2a    3c 

.        9 ex    4ab  •,  6acd 

denommator.                                     ^^^*-  -r—'  3— >  ^^^^  -5 — 

oac    6ac  oac 

3  2a;  2x 

8.  Reduce  _,  — ,  and  «+ — ,  to  fractions  having  a  com- 

4  3  a 

1  .     ,  .         9a     Qax         .   \2a^-\-24x 

mon  denommator.  Ans. ,  — ,  and ^ . 

12a    12a  12a 


B2  ELEMENTARY  ALGEBRA. 

(Art.  43.)  The  preceding  rules  are  general,  and  corres- 
pond to  quantities  that  are  prime  to  each  other ;  but  in  cases 
of  multiple  denominators,  the  general  rule  would  carry  the 
operator  through  a  much  longer  process  than  necessary. 

We  will,  therefore,  investigate  a  more  convenient  practical 
rule,  which  will  apply  to  fractions  having  multiple  denomina- 
tors.    For  example,  we  require  the  sum  of  the  fractions 
a      c       d 
b     nh     nib 

As  before,  we  designate  the  sum  by  S,  which  gives  the 
equation  , 

Multiplying  every  term  by  h,  then  we  have 

hS=a-\--~{-^  (2) 

n     m 

Multiplying  by  n,  and  then  by  m,  or  multiply  at  once  by 
WW,  then  we  have 

nmhS=anm-]rcm-{-dn  (3) 

Dividing  equation  (3)  by  nmh,  and  we  have 

anm-\-c'm-\-dn 
nmb 
The  product  nmb  is  composed  of  all  the  different  factors  in 
the  denominators,  and  no  more  ;  it  is,  therefore,  the  least  com- 
mon multiple  of  the  denominators,  (Art.  32). 

To  find  the  numerators,  we  divide  this  product  by  the  de- 
nominator of  any  one  of  the  fractions,  and  multiply  the  quotient 

by  the  numerator.     For  instance,  take  the  first  fraction,  _. 
^  b 

Divide  nmb  by  b,  and  we  have  nm ;  multiply  this  by  a,  and 
we  have  anm,  the  new  numerator  for  the  first  fraction ;  and 
by  the  same  operation  we  find  the  numerators  for  the  other 
fractions. 


ADDITION  OF  FRACTIONS.  83 

Hence,  we  have  the  following  rule  for  reducing  fractions  to 
equivalent  fractions  having  a  least  common  denominator,  and 
thence  finding  their  sum. 

R  u  L  E  1 . — Find  the  least  cmnmon  multiple  of  all  th.e  denom- 
inators, which  will  he  the  least  common  denominator. 

2.  Divide  the  common  denominator  hy  the  denominator  of  the 
first  given  fraction,  and  multiply  the  quotient  hy  the  numerator, 
the  product  will  he  the  first  of  the  required  numerators. 

3.  Proceed  in  like  manner  to  find  each  of  the  required 
numerators. 

4.  The  sum  of  the  fractions  will  he  the  algehraic  sum  of 
these  numerators,  with  the  common  denominator  under  them. 

Note  . — The  fractions  should  be  reduced  to  their  lowest 
terms  before  this  or  the  preceding  rules  are  applied. 

OTHER    EXAMPLES. 
1.  Add  i, -|,  |,  and  y^  together.  .     .     .     Ans.  ^=-1^. 

The  least  common  multiple  of  these  denominators  is  obvi- 
ously 12;  therefore,  multiply  both  members  of  the  equation 
by  12,  and  we  have 

125=6-{-8+10-{-7  (2) 

Dividing  again  by  12,  and  we  have 

'S'=/^+A+1|+t\  (3) 
The  second  member  of  equation  (3)  is  composed  of  equiv- 
alent fractions  to  those  in  equation  (1),  as  may  be  seen  by 
reducing  these  fractions  to  their  lowest  terms.-  In  equation 
(3),  the  fractions  have  a  common  denominator,  composed  of 
the  least  common  multiple  of  the  original  denominators. 
The  sum  of  these  fractions  is,  of  course,  the  sum  of  the 


84         ..  ELEMENTARY  ALGEBRA. 

numerators  with  tlie  common  denominator  under  it ;  tliua,  f  |, 
and  might  have  been  taken  for  equation  (2),  thus  : 

12>S'=31 
Or S=^ 

2.  A.dd ,  and  together.    .     .     Ans.  — ! — 

35c  7c         ^  5c 

3.  Add  ,  — ^,  and     ^     together.    .      Ans. 

1-fa    i_-a  i+a      ^  l_a 

4.  Add  -,  — ,  and  —  together.    .     .     Ans. 

b    3b  4a      ^  I2ab 

5.  Add ! ,  and  tosfether. 

125c  35         ° 

Ans. 

4c 

Note. — Examples  2,  3,  and  5,  and  all  others  like  them, 
had  better  be  performed  by  solving  an  equation.  If  not  so 
performed,  multiply  the  numerator  and  denominator  of  the 
second  fraction  in  example  2,  by  5,  and  in  example  5,  by  4c, 
and  thus  make  the  denominators  common  by  inspection.  Then 
unite  the  numerators,  and  reduce  to  lowest  terms. 

In  examples  like  the  following,  consisting  of  entire  quanti- 
ties and  fractions,  make  two  examples  of  the  operation,  by 
first  uniting  the  entire  quantities,  and  then  the  fractions,  and 
lastly  uniting  the  two  sums  together  by  their  proper  signs. 

6.  Add  2x,  Sxj^ —  and  a;_j together.  Ans.  6a:_j 

7.  Add  5xj^ ZL  and  4ar__ — ~~  together. 


•  15a; 

8.  Add and  together.     Ans. . 

(^On-bXa+b)  a+5      ^  a— 5 


ADDITION  OF  FRACTIONS.  85 

9.  Add  ,  -HI   and  JH^  together.       ,     .    Ans.  0 

ab        be  ac 

10.  Add  "-— ^  and  ?II^  together.    .     .     .    Ans.  ?IJ? 

ax  X  a 

11.  Add  5±5,  5=^  and  1  together. 

y  ay  3a  15a4-5y+9 


12. 


Add  ^^  and  ^  together.      .    ,  Am.  ^1^11 


a~-6  a+6     °  a'—b^ 

13.  Add  -,  and together. 

b       cd  bed  J     Aii  \    9 

Am.  '!±:^±^. 


bed 

14.  Add  _fL  and  -1-  together.     .    .     .   Ans.  ^H' 
a-{.h  a-^      °  a2_^» 

16.  Add  and  — ^—„  together.  .     .     .  Ans.  — ^-, 

16.  Add  J^  and  ln?-  together.  .     .     .   Ans.  ^"*"'^* 
l__a4  l+a2 


1+a'      °  1— a' 

17.  Add  1+1  and  1— (^)   together.  .     .     An^,  1 


SUBTRACTION    OF    FRACTIONS. 

(Art.  44.)  We  would  remind  the  pupil  that,  in  addition, 
we  took  the  sum  of  the  numerators,  after  the  fractions  were 
reduced  to  a  common  denominator.  Hence,  the  difference  of 
the  two  fractions  must  be  found  by  taking  the  difference  of 
their  numerators,  when  the  denominators  are  alike.  For  ex- 
ample, the  difference  between  /^  and  /j,  must  be  fV=i>  and 


86  ELEMENTARY   ALGEBRA. 

the  difference  between  |  and  f ,  must  be  f ,  &c.  These  obser- 
vations must  give  us  the  following 

Rule  1 . — Reduce  the  fractions  to  a  common  denominator. 

2.  Subtract  the  numerator  of  the  subtrahend  frcm  the  numer- 
ator of  the  minuend,  and  place  the  difference  over  the  commrni 
denominaior. 

(Art.  45.)  We  may  also  find  the  result  of  any  proposed 
example  by  means  of  equations,  as  in  addition. 

For  example,  from  |  take  -f.  The  remainder  is  some  num- 
ber, which  we  may  represent  by  R. 

Then        -^=f--f         (1) 

Or,  we  may  consider  that  in  every  possible  example,  the 
remainder  and  subtrahend  added  together,  must  equal  the 
minuend ;  that  is* 

Ii+^=i  (2) 

Equation  (2)  is  the  same  as  equation  (1),  except  the  frac- 
tion ^  is  transposed,  according  to  the  rule  of  transposition 
on  page  1 7. 

Multiply  equation  ( 1 )  by  7,  and  we  have 
7i2=V— 5 

Multiplying  by  4,  and  28i2=2 1—20=1 

By  division  .     ,     ,     .     B=-^^ 

EXAMPLES, 


1.  From  !^  take  ^-^l  An.,  21^Z^f±?_l!^: 

2                3  6          ~"      6 

2.  From  _L  take  — .  Eq.  fractions    ^"^^ ,  ^t. 

x—y  ■         x-\-y  ^—y^    ^—^^ 


Difference  or  Ans. "-^    . 


*  We  take  this  view  of  the  subject  to  show  the  pupil  the  nature  of 
equations ;  not  that  it  is,  or  is  not,  a  better  method  of  solving  the 
problems. 


SUBTRACTION  OF  FRACTIONS.  87- 

3.  From  -  take  —., Difif.  — 

4.  From  ?e^  take  ^.    .     ..'...      ^Ans.  ^H^ 

3  2,  6 

5.  From  take  — IZ — Ans. 

a-l-1  a2— a+1  l+a^ 

6.  From  —  take  — Ans. 


4a  2x  4ax 


7.  From  —  take  ^- 

4x.           3a 

12ax 

8.  From    ^'''^^  take     «'*^   .... 

4a2— 5^            2a+b^ 

A                 0,'b' 

.     Ans. 

4a^—b' 

OPERATION. 

39  J . 

2a-\-b'^~(2a+b^)(2a—b') 

{4a^—b')R-^2a^b''  -<i%^=2a%^ 

Dropping  from  both  members  (2a^5),  and  transposing  a^5*, 
and  we  have 

{4a^—b')R=a^b'' 

d'b'^ 


By  division,        ....     jR= 


4d'—b' 


9.  From  —  take  -1_ ^n*.  tz^. 

x—\  x-\-\  s^—1 

10.  From  2a—2x-{-^.II^  take  2a— 4a:+?II^. 

Ans.  2ar-f 

n.  From  t±^  fake  ^fzf.      ....      Ans.  1?^. 
5c  7c  35ff 

12.  From  ^±1  take  !if±?.        .  Am.  ( -lUfZll^ 


ELEMENTARY  ALGEBRA. 


13.  From  ^i:^  take  ^.        .     .     .    Ans.  ( ^^11^) 

14.  From  1±5_'  take  izi^' Ans,     ^^' 


1—d'  1+a*  l—a* 

15.  From  x-^^Zl-  take  -f^.      .     .  Arts.  x-J^. 

:^-\-xy  ar — xy  ar — if 

16.  From  ?=?  take  5^r±«.    ^m.  Sf*z:Si*-4Jc+8ac 

2(;  5rf  lOcrf 

17.  From  !(?!±?1)  take  ?=? ^n..  ?+*. 

a^ — 6^  a-t-6  a — 1> 

18.  From  JL.  take  ^ ^W5.       ^ 


a; — 3  a;  ar — 3a; 

19.  From  6a+ll^^=H  take  Aa^^^±. 

^w*.  2a+^±r. 
6 


SECTION   II 


EQUATIONS. 

The  most  interesting  and  the  most  essential  part  of  Alge- 
bra is  comprised  in  equations  ;  and  nearly  all  of  our  previous 
preparations  have  been  with  a  view  to  a  more  ready  under- 
standing of  equations. 

The  use  of  equations  is  the  solving  of  problems  in  almost 
every  branch  of  mathematical  science,  and  also  in  the  inves- 
tigation of  scientific  truths. 

For  instance,  we  have  already  investigated  rules  for  the 
addition,  subtraction,  and  division  of  fractional  quantities, 
by  means  of  algebraical  equations,  and  we  have  thus,  inci- 
dentally, given  some  explanations  concerning  the  nature  of 
equations  ;  but  now  coming  to  the  subject  in  order,  we  shall 
disregard  all  this,  and  commence  on  the  supposition  that  the 
pupil  must  yet  learn  every  particular. 

(Art.  46.)  An  equation  is  an  algebraical  expression,  mean- 
ing that  certain  quantities  are  equal  to  certain  other  quantities. 
Thus,  34-4=7;  a-\-b=G;  x-\-4  =  \0,  are  equations,  and 
express  that  3  added  to  4  is  equal  to  7,  and  in  the  second 
equation,  that  a  added  to  b  is  equal  to  c,  &c.  The  signs  are 
only  abbreviations  for  words. 

The  quantities  on  each  side  of  the  sign  of  equality  are 
called  members.  Those  on  the  left  of  the  sign  form  the  Jirst 
member,  those  on  the  right,  the  second  member. 


90  ELEMENTARY  ALGEBRA. 

(Art.  47.)  As  unlike  things  can  neither  be  added  to,  nor 
subtracted  from  each  other,  it  follows  that  a  member  of  an 
equation  must  consist  of  the  same  kind  of  quantities  ;  and  as 
it  is  absurd  to  suppose  one  kind  of  quantity  equal  to  another 
in  any  other  sense  than  a  numerical  one,  it  also  follows  that  the 
members  of  an  equation  must  be  equal  in  kind  as  well  as  in 
number.     That  is.       Dollars  =  Dollars, 

Or        ....     Pounds  =  Pounds,  &c.,  &c. 

It  is  true  we  may  say  that  a  farmer  has  as  many  dollars 
in  his  purse  as  he  has  sheep  and  cows  on  his  farm. 

Here  we  cannot  say  that  his  sheep  and  cows  are  eqval  to 
his  dollars  ;  but  the  number  of  his  sheep  added  to  the  number 
of  his  cows,  are  equal  to  the  number  of  his  dollars. 

That  is,     .     .      Kumber  =  Number. 

Indeed,  when  dollars  equal  dollars,  or  yards  equal  yards, 
it  is  but  really  a  number  of  dollars  equal  to  a  number  of 
dollars,  cfec;  that  is,  universally,  number  equal  to  number. 

(Art.  48.)  In  the  solution  of  problems,  every  equation  is 
supposed  to  contain  at  least  one  unknown  quantity ;  and  the 
solution  of  an  equation  is  the  art  of  changing  and  operating 
on  the  terms  by  means  of  addition,  subtraction,  multiplica- 
tion, or  division,  or  by  all  these  combined,  so  that  the  un- 
known quantity  may  stand  alone  as  one  member  of  the  equa- 
tion, equal  to  known  quantities  in  the  other  member,  by  which 
it  then  becomes  known. 

Every  equation  is  to  be  regarded  as  the  statement,  in  alge- 
braic language,  of  a  particular  question. 

Thus,  X — 3=4,  may  be  regarded  as  the  statement  of  the 
following  question:  To  find  a  number  from  which,  if  3  be 
subtracted,  the  remainder  wuU  be  equal  to  4. 

An  equation  is  said  to  be  verified,  when  the  value  of  the 
unknown  quantity  being  substituted  for  it,  the  two  members 
are  rendered  equal  to  each  other. 


EQUATIONS.  91 

Thus,  in  the  equation  x — 3=4,  if  7,  the  value  of  a?,  be 
substituted  instead  of  it,  we  h^ve  7 — 3=4, 
Or 4=4. 

(Art.  49.)  Equations  are  of  the  fir  sty  second,  third  and 
higher  degrees,  according  to  the  highest  power  of  the  unknown 
quantity  involved. 

'     _i_z  ~     !•  are  equations  of  the  first  degree. 

,T ,  _T    >■  are  equations  of  the  second  degree. 

^     9  ,  ,  __     [-  are  equations  of  the  third  degree,  <fec. 

Equations  of  the  first  degree  are  also  called  simple  equa- 
tions, and  equations  of  the  second  degree  are  called  quadratic 
equations ;  but  quadratic  equations  may  include  many  other 
equations  of  any  even  degree,  according  to  certain  relations 
that  may  exist  between  the  several  parts  of  the  equation, 
which  will  be  explained  hereafter.  At  present  we  shall  con- 
fine our  investigations  to  simple  equations. 

(Art.  60.)  Equations  are  either  numeral  or  literal.  Nu- 
meral equations  contain  numbers  only,  excepting  the  un- 
known quantity.  In  literal  equations,  the  given  quantities 
are  represented  by  letters,  in  whole  or  in  part. 

An  identical  equation,  is  one  in  which  the  two  members  are 
identical ;  or,  one  in  which  one  of  the  members  is  the  result 
of  the  operations  indicated  in  the  other. 

Thus,  2r— l=2a;— 1   )  -a     ,•     ^  +• 

6^+3.:=  8a:         [  are  identical  equations. 

(Art.  51.)  The  unknown  quantity  of  an  equation  may  be 
united  to  known  quantities,  in  four  different  ways  ;  by  addi- 
tion, by  subtraction,  by  multiplication,  and  by  division,  and 
further  by  various  combinations  of  these  four  ways,  as  shown 
by  the  foUowiog  equations,  both  numeral  and  literal : 


ELEMENTARY  ALGEBRA. 


NUMERAL. 

LITERAL. 

1st. 

By  addition,      .     . 

.  ^a;+6=10 

x+a=b 

2d. 

By  subtraction, 

.     a:— 8=12 

X—C=d 

3d. 

By  multiplication,  . 

.       20a:=80 

ax=e 

A  4.1. 

■D-,  J :,„•„: 

X 

X 

4tli.  By  division _16  5=^+« 

5th.  a;+6— 8+4=10-1-2— 3,  x-\-a—h-hc=d-\-c,  &c.,  are 
equations  in  which  the  unknown  is  connected  with  known 
quantities,  both  by  addition  and  subtraction. 

2a;+-=21,    ax-\--=^c,  are  equations  in  which  the  unknown 
3  0 

is  connected  with  known  quantities,  by  both  multiplication  and 

division. 

Equations  often  occur,  in  solving  problems,  in  which  all  of 
these  operations  are  combined. 

(Art.  52.)  Let  us  now  examine  and  discover,  if  possible, 
how  the  unknown  quantity  can  be  separated  from  known 
quantities,  and  be  made  to  stand  alone  as  one  member  of  the 
equation.     For  this  purpose,  let  us  take  the  equation 

x-\-a=^h 
Take  equals  from  equals,     a=a 


Remainders  are  equal,         x=^h — a     (AlX.  2). 

Here  the  quantity  a,  connected  to  x,  appears  on  the  other 
side  of  the  equation,  with  its  oj^posite  sign. 

Again,  suppose  we  have  the  equation 

a;— 8=10 
Add  equals  to  equals,  8=8 

Sums  will  be  equal,  a:=104-8     (Ax.  1). 

Here,  again,  the  quantity  connected  with  x  appears  on  the 
opposite  side  of  the  equation,  with  its  opposite  sign.  , 


EQUATIONS.  $f 

From  this  we  derive  tlie  following  operation,  which  operation 
is  called 

TRANSPOSITION. 

Rule. — Any  quantity  may  he  changed  from  one  memler  of 
an  equation  to  the  other,  if,  in  so  doing,  we  change  its  sign. 

Now,  suppose  we  have  an  equation  in  the  form  of 
ax=c 

Here,  x  is  united  to  a  by  multiplication ;  it  can  be  disunited 
by  division.     Dividing  by  a,  gives 

c 

x=- 
a 

Again,  suppose  an  equation  appears  in  the  form  of 

X 

Here,  x  is  united  to  a  known  quantity  by  division,  and  it 
can  be  disunited  by  multiplication  ;  that  is,  multiply  by  a,  and 
we  have  x=^ag 

From  these  observations,  we  deduce  this  general  principle : 

That  to  separate  the  unknown  quantity  from  additional  tei^mSy 
we  must  use  sid^traction ;  from  subtracted  terms,  we  must  use 
addition;  from  multiplied  terms,  we  must  use  division;  from 
division,  we  must  use  multiplication. 

In  all  cases  take  the  opposite  operation. 

(Art.  53.)  In  many  practical  problems,  the  unknown 
quantity  is  often  combined  with  the  known  quantities,  not 
merely  in  a  simple  manner,  but  under  various  fractional  and 
compound  forms.  Hence,  rules  can  only  embody  general 
principles,  and  skill  and  tact  must  be  acquired  by  close  atten- 
tion and  practical  application  ;  but  from  the  foregoing  princi- 
ples, we  derive  the  following 

General  Rule  . — Connect  and  unite,  as  much  as  possi- 
ble, all  the  terms  of  a  similar  kind  on  both  sides  of  the  equation. 


li  ELEMENTARY  ALGEBRA. 

Then,  Id  clear  of  fractions,  multiply  both  sides  by  the  denomina- 
lors,  erne  after  another,  in  succession.  Or,  multiply  by  their 
continued  product,  or  by  their  least  common  multiple,  (when 
such  a  number  is  obvious),  and  the  equation  will  be  free  of 
fractions. 

Then  transpose  the  unknown  terms  to  the  first  memher  of  the 
equation,  and  the  known  terms  to  the  other.  Then  unite  the  sim- 
ilar term^,  and  divide  by  the  coefficient  of  the  unknoum  term,  and 
ike  equation  is  solved. 

EXAMPLES. 

1.  Given  3x — 2-\-5=2x-\-l2,  to  find  x.      .     Ans.  xz=9. 
By  transposition,     Sx — 2ar=12+2 — 5 

Unitino:  terms,  x=9 

In  place  of  transposing,  we  may  drop  equals  from  both 
Bides,  or  add  equals  to  both  sides,  as  the  circumstances  may 
require. 

In  the  present  example,  we  d7vp  2x  from  both  sides,  and 
conceive  — S-f-S  united,  then  we  have 
x-\-3=n 

Drop  3  from  both  sides,  and  we  have 

x—9,  as  before. 

Dropping  and  transposing  is  one  and  the  same  operation, 
differing  only  in  form. 

2.  Given  6— 2a;+10=:20— 3a;— 2,  to  find  a:.     Ans.  a.-=2. 
Uniting  similar  terms  in  both  members,  we  have 

.—2a;-}- 16  =  18— 3a: 
Adding  3a;  to  both  sides,  and  dropping  16  from  both,  we 
have  a;=2 

3.  Given  |__^-|-a;=9a;— 3,  to  find  x.      .     .     Ans.  x^4. 
Drop  X  from  both  members,  then  we  liave 

X      X 


Arts. 

x=  3. 

Am. 

x==  5. 

Ans. 

x=  9. 

Arts. 

a;=12. 

Ans. 

x=^  m. 

EQUATIONS.  95 

Multiply  every  term  by  4,  and  we  have 

2;r— a:~4a; — 12 
Transpose  Ax^  and  unite. 
Then       ....     — 3a;=  —12 
Divide  both  members  by  — 3,  and  x=A. 

4.  Given  5a;+22 — 2a;==31,  to  find  x.    . 

5.  Given  4a;4-20— 6=34,  to  find  x.     . 

6.  Given  3a;-l-12-|-7a;=102,  to  find  ar. 

7.  Given  10a;— 6a;-}- 14=  62,  to  find  ar. 

8.  Given  ax-\rbx='ma-]r'^i,  to  find  x.   . 
Separate  both  members  into  their  prime  factors. 
Thus,  ....     {a-{-h)x={a-{-h)m. 
Dividing  both  members  by  (a+J),  gives  x=m. 

a — c 

9.  Given  ax-\-dx=a — c,  to  find  x.    .     .      Ans.  x==--j^,' 

10.  Given  3(a;-{-l)-i-4(a;-f2)=6(a;+3),  to  find  x. 

Ans.  a;=7. 
Perform  the  multiplication  indicated,  then  reduce. 

Sa*  X     X 

11.  Given  — 4- 1 6=-+ -+17,  to  find  ar.      .     Ans.  a;=8. 

In  the  first  place,  drop  16  from  both  members,  according 
to  the  general  rule.     Then 

3a;    2;    a; 
T"^2"^8"^^ 

Multiply  both  members  by  8,  the  least  common  multiple 
of  the  denominators,  and  we  have 

,  6a;=4a;+a;-|-8,  or,  a;=8 

12.  Given  5--3-{--=5— 3,  to  find  x.  .     .     Ans.  a;=  6. 

2  3 

13.  Given  ?— ?+2=3,  to  find  a-.    .     .     .     Ans.  a;=12. 

3  4 


96  ELEMENTARY   ALGEBRA. 

14.  Given  ? 4- -_-_-,  to  find  a;.  .     .     .     Ans,    x=2. 

4^8     6~12 

5x     1     11     7x 

15.  Given  "8'~^4~"6"~^T2*  *^  ^^^  ^'     *     '     ^^'  ^~^^* 

a;     X — 5  2ab-\-5a 

16.  Given  --1-— ^+25=3^,  to  find  iP.  ^W5.  a;=— ^x 

3x  X 

17.  Given  y-l-2j4-n=^+17,  to  find  x.      Ans.  a;=10. 

18.  Given  ^x~{-ix-\-^x=S9,  to  find  the  value  of  x. 

Here  are  no  scattering  terms  to  collect,  and  clearing  of 
fractions  is  the  first  operation. 

By  examination  of  the  denominators,  12  is  obviously  their 
least  common  multiple,  therefore,  multiply  by  12. 

Hence,     .     .     .     Gx-{-4x+3x=39X12 
Collect  the  terms,  13a;=39  X  12 

Divide  by  13,  and  x=  3X12=36,  Ans. 

19.  Given  |a;+-Ja;+ia;=a,  to  find  x. 

This  example  is  essentially  the  same  as  the  last.  It  is 
identical  if  we  suppose  a=39. 

Solution,  .     .     .     6x-]-4x+3x=l2a 

Or, 13a;=12a 

12a 
Divide  and ^"13" 

Kow  if  a  be  any  multiple  of  13,  the  problem  is  easy  and 
brief  in  numerals. 

20.  Given  irr— 5-f ia;+8+ia;— 10=100— 6— 7  to  find 
the  value  of  x. 

Collecting  and  uniting  the  numeral  quantities,  we  have 
ia:+{«+i(r=94 


f 


EQUATIONS.  97 

Multiply  every  term  by  60,  and  we  have 

20x-{-\5x-]-nx==94'60 
Collecting  terms,  47a:=94*60 

Divide  both  sides  by  47,  and      a?=  2-60=120,  Ans. 

21.  Given  ia;-|-|a;-{-ia;-l-ia;=  77,  to  find  ar.  Ans,  x=:  60. 

22.  Given  ia:-l-ia;-|-ia:=130,  to  finda;.     .  Ans.  x=120, 

23.  Given  ^x-\-}z-{-j'jX=90,  to  find  x.     .  Ans.  a:=120. 

24.  Given  iy+|y+4y=82,  to  find  y.       .  Ans.  y=  84. 

25.  Given  5x~\-^x-r-^x=34,  to  find  x. .     .  Ans.  x=     6. 

N.  B.  In  solving  21,  22,  23,  24  and  25,  take  19  for  a 
model,  and  write  a  to  represent  the  second  members  of  the 
equations,  to  save  numeral  multiplications. 

3a?     a:— 1  20a:-fl3 

26.  Given  -j — —^  =6a; — r to  find  x. 

Multiply  by  4,  to  clear  of  fractions,  and 

3a: — 2x-\-2=24x — 20a; — 13.     Reduced,  x=5. 

(Art.  54.)  When  a  minus  sign  stands  before  a  compound 
quantity,  it  indicates  that  the  whole  is  to  be  subtracted ;  but 
we  subtract  by  changing  signs,  (Art.  5).     The  minus  sign 

before  in  the  last  example,  does  not  indicate  that  the  x 

2 

is  minus,  but  that  this  term  must  be  subtracted.     When  the 

term  is  multiplied  by  4,  the  numerator  becomes  2a; — 2,  and 

subtracting  it,  we  have  — 2x-\-2. 

Q. 3     9     a;-l-4 

27.  Given  x — ~o~=q — ~3~*  *°  ^^^  ^*  '     ^^*  ^~  ^' 

J.  I    O       />«       g  X—  1 

28.  Given  — g— — ^-+2=3;— — g-,  to  find  x. 

Ans.  x=  7. 

29.  Given  ^•^"~^_,^^~^=1,  to  find  a:.      .     Ans.  ar=  6. 

11  13 

30.  Given  ^ — —^=—54-0-,  to  find  a;.  .     Ans.  x=]0. 


98  ELEMENTARY  ALGEBRA. 


PROPORTION. 

(Art.  65.)  Sometimes  an  equation  may  arise,  or  a  prob- 
lem must  be  solved  through  the  aid  of  proportion. 

Proportion  is  nothing  more  than  an  assumption  that  the 
same  relation^  or  the  same  ratio  exists  between  two  quantities 
as  exists  between  two  other  quantities. 

Quantities  can  only  be  compared  when  they  are  alike  in 
kind,  and  one  of  them  must  be  the  unit  of  measure  for  the 
other. 

Thus,  if  we  compare  A  and  ^,  we  find  how  many  times y 

or  part  of  a  time,  A  is  contained  in  By  by  dividing  B  by  Ay 

thus,  B 

-j=r,    or    l>=rA 

That  is,  a  certain  number  of  times  A  is  equal  to  B. 

Now  if  we  have  two  other  quantities,  C  and  J),  having  the 
same  relation  or  ratio  as  A  to  B,  that  is,  if  JD=rO, 

Then  ^  is  to  -B  as  0  is  to  J). 

But  in  place  of  writing  the  words  between  the  letters,  we 
write  the  signs  that  indicate   them. 

Thus,     .    .     .     .     A:B::C:J) 

But  in  place  of  B  and  J),  write  their  values  rA,  and  r(7. 

Then,     .     .     .     .      A:rA::C:rO 

Multiply  the  extreme  terms,  and  we  have  rCA. 

Multiply  the  mean  terms,  and  we  have  rA  0. 

Obviously  the  same  product,  whatever  quantities  may  be 
represented  by  either  A,  or  r,  or  C. 

Hence,  to  convert  a  proportion  into  an  equation,  we  have 
the  followinor 

o 

Rule  . — Place  the  produd  of  the  extremes  equal  to  the  pro^ 
duct  of  i/ie  means. 


EQUATIONS.  99 

(Art.  56.)  The  relation  between  two  quantities  is  not 
changed  by  multiplying  or  dividing  both  of  them  by  the 
same  quantity.  Thus,  a:b:  :2a:2b,  or  more  generally, 
a:b:  :na:nh,  for  the  product  of  the  extremes  is  obviously 
equal  to  the  product  of  the  means. 

That  is,  a  is  to  5  as  any  number  of  times  a  is  to  the  same 
number  of  times  b. 

We  shall  take  up  proportion  again,  but  Articles  55  and  56 
are  sufficient  for  our  present  purpose. 

EXAMPLES. 

\.  If  2>  'pounds  of  coffee  cost  25  ce7iis,  what  will  a  bag  of  60 
pounds  cost  ?  Ans.  500  cents. 

Ans.  It  will  cost  a  certain  number  of  cents,  which  I  desig- 
nate by  X,  and  the  numerical  value  of  x  can  be  deduced  from 
the  following  proportion  :  Pounds  compare  with  pounds,  as 
cents  compare  with  cents.  That  is,  these  different  kinds  of 
quantities  must  have  the  same  numerical  ratio. 

Thus,      ....     3:G0::25:x 

Without  the  x,  this  is  the  rule  of  three  in  Arithmetic,  be- 
cause there  are  three  terms  given  to  jSnd  the  fourth ;  and  in 
Algebra  we  designate  the  fourth  term  by  a  symbol  before 
we  know  its  numerical  value,  which  makes  the  proportion 
complete. 

By  the  rule  (Art.  55),  3a;=60*25 

60*25 


Or, 


3 

Hence,  when  the  first  three  terms  of  a  proportion  are  given 
to  find  the  fourth,  multijdy  the  second  and  third  together,  and 
divide  by  the  first. 

In  Arithmetic  it  requires  more  care  to  state  a  question  than 
it  does  in  Algebra,  because  in  the  former  science  we  have  not 
so  much  capital  at  command  as  in  the  latter. 

In  Algebra  it  is  immaterial  what  position  the  nidinown  term 


100  ELEMENTARY  ALGEBRA. 

has  in  the  proportion,  if  the  comparison  is  properly  made. 
Thus,  in  the  foregoing  question  the  demand  is  money,  and 
money  must  be  compared  with  money  ;  and  the  statement  may 
be  made  thus,    .     .     25 :  a; :  :  3  :  60 

Or  thus,    .     .     .     ic :  25  : :  60  :  3 

From  either  one  of  these  proportions  the  value  of  x  is  found 
by  multiplying  and  dividing  by  the  same  numbers. 

2.  If  2  cords  of  wood  cost  5  dollars,  what  will  48  cords 
cost?  Ans.  ^120. 

Given  5  :  a; :  :  2  :  48,  to  find  x. 

3.  Given  2:x:  :  6  :  5x — 4,  to  find  x.       .     .     Ans.  x=2. 
The  equation,    -.     .     10a: — 8=6a; 

4.  Given  (li::}}(^l} :  ^^  :  2a: :  1,  to  find  x. 

ScT^  3a 

Divide  the  first  two  terms  by  (x-{-l),  (Art.  56).  Also 
multiply  by  3a. 

6.  Given  x-\-2  :a:  :b:c,  to  find  the  value  of  x. 

Ans.  x=— — 2. 
c 

6.  Given  2a; — 3 :  x — 1  :  :  2a; :  a:-f  1,  to  find  the  value  of  x. 

Ans.     x=3. 

7.  Given  x-\-6  :  38 — x  :  :  9  :  2,  to  find  x.         Ans.  a;=30. 

8.  Given  x-^4  :  x—1 1  :  :  100  :  40,  to  find  x.   Ans.  x=2\. 

9.  Given  x-\-a : x — a :  :c:d,  to  find  x.   Ans.  x=  ^        1. 

c — d 

10.  Given  a;  :2a; — a:  :a  :5,  to  find  a;.     .     Ans.  x=  — 

^a—h 

1\.  Given  a :b:  :2y:d,  to  ^ndy.     .     .     .      Ans.  y—-   ' 

26. 

12.  Given  a^ — ac  '.ax\:\:  (d—b),  to  find  x. 

Ans.  x=(d-'b){a—<). 

18.  Given  x :  76 — x :  :  3  : 2,  to  find  x.  .     .      Ans.  x=45. 


EQUATIONS.  101 

/. 
QUESTIONS  PRODUCING  SIMPLE   EQUATIONS. 

(Art.  57.)  We  now  suppose  flie  p,i?p|l '^can  readily' t^duce 
a  simple  equation  containing  Ji^tone  iinknbwn  qua'ntfty,  and 
he  is,  therefore,  prepared  to  ^§ol\t  .the;  j©libWii^_;qiiefeti^ns. 
The  only  difficulty  he  can  experience,  is  the  want' of  tact  to 
reason  briefly  and  poAverfully  with  algebraic  symbols ;  but 
this  tact  can  only  be  acquired  by  practice  and  strict  attention 
to  the  solution  of  questions.  We  can  only  give  the  following 
general  direction : 

Represent  the  unknovm  quantUy  hy  some  symhU  or  letter,  and 
really  consider  it  as  definite  and  known,  and  go  over  the  same 
operations  as  to  verify  tlie  answer  when  known. 

EXAMPLES. 

1.  A  merchant  paid  $480  to  two  men,  A  and  B,  and  he 
paid  three  times  as  much  to  B  as  to  A.  How  many  dollars 
did  he  pay  to  each  ?  Ans.  To  A,  $120,  to  B,  $360. 

Let ar=  the  sum  to  A, 

Then  ......    3a:=  the  sum  to  B, 

Sum Ax=  the  sum  paid  to  hoth. 

But 4Q0=  the  sicm  paid  to  hoth. 

Thus,  when  any  question  has  been  clearly  and  fully  stated, 
it  will  be  found  that  some  oondition  has  been  represented  in 
two  ways ;  one  having  the  unknovm  quantity  in  it,  and  the 
other  having  a  known  quantity.  These  two  expressions  must 
be  put  in  the  same  line,  with  the  sign  =  between  them,  so  as 
to  form  an  equation.  And  then,  by  reducing  the  equation, 
the  required  result  will  be  found. 

Thus,     .     .     4a;= 480,  therefore,  a;=  120. 

Observe  that  the  problem  would  be  essentially  the  same, 
whatever  number  of  dollars  were  paid  out.  It  is  not  n^es- 
sary  that  the  number  should  have  been  48C,  any  more  than 


102  ELEMENTARY  ALGEBRA. 

48,  or  any  otlier  number.     Therefore,  to  make  the  problem 
more  general,  we  may  represent  the  number  of  dollars  paid 

out  ti^av^d  the  6Cftia<|o^  yiiiW  then  be  4a;=a.     And  a;=-. 

Again,  the  problem  w-9u>d  h^-ve  been  the  same  in  charac- 
ter,.'an  J  "eqlfaBy -as  "stmf^leVhjJd- the  merchant  paid  4  times, 
or  5  times,  or  n  times  as  much  to  B  as  to  A. 

We  may  therefore  make  it  general  by  stating  it  in  the 
following  words : 

A  merchant  paid  a  dollars  to  two  men,  A  and  B,  and  he  paid 
n  times  as  many  dollars  to  B  as  to  A.  What  did  he  pay  to 
each? 

a;=  the  sum  to  A, 
nx^=  the  sum  to  B, 


Let      . 
Then  . 
By  add. 
Or      . 
Also,  . 

Therefore 


.    x-\-nx 

(]+w)rr=  the  sum  to  both. 
a=  the  sum  to  both. 

Cl"i~wV=«>  or  a:= 

l4-7^• 


This  shows  that  the  sum  paid  to  A  was dollars,  and 

^  \-\-n 

as  B  had  n  times  as  many,  the  sum  to  B  was . 

^  \^n 

For  proof,  4~ must  equal  a.     As  the  denomina- 

,  ,1  X?  ii     X       •    a-\-7ia         (\4-n)a 

tors  are  common,  the  simi  of  the  two  is  — i —    or  i — ! — I 

or  a,  by  suppressing  the  common  factors  in  numerator  and 
denominator 

2.  My  horse  and  saddle  are  worth  $100,  and  my  horse  is 
worth  7  times  my  saddle.     What  is  the  value  of  each  ? 

Ans.  Saddle,  ^12^;  horse,  887-^^. 


EQUATIONS.  108 

3.  My  horse  and  saddle  are  worth  a  dollars,  and  my  horse 
IS  worth  n  times  my  saddle.     What  is  the  value  of  each  ? 

Ans.  Saddle,  _^  ;  horse,  ^^. 

4.  A  farmer  said  he  had  4  times  as  many  cows  as  horses, 
and  5  times  as  many  sheep  as  cows ;  and  the  nimiber  of 
them  all  was  100.     How  many  horses  had  he  ?  Ans.  4. 

5.  A  farmer  said  he  had  n  times  as  many  cows  as  horses, 
and  m  times  as  many  sheep  as  cows ;  and  the  number  of 
them  all  was  a.     How  many  horses  had  he  ? 

Ans. horses. 

\-\-n-\-mn 

6.  A  school-girl  said  that  she  had  120  pins  and  needles; 
and  that  she  had  seven  times  as  many  pins  as  needles.  How 
many  had  she  of  each  sort  ?     Ans.  15  needles,  and  105  pins. 

7.  A  teacher  said  that  her  school  consisted  of  64  scholars ; 
and  that  there  were  three  times  as  many  in  Arithmetic  as  in 
Algebra,  and  four  times  as  many  in  Grammar  as  in  Arithme- 
tic.    How  many  were  there  in  each  study  ? 

Ans.  4  in  Algebra ;  12  in  Arithmetic  ;  and  48  in  Grammar. 

8.  A  certain  school  consisted  of  a  number  of  scholars  ;  a 
certain  portion  of  them  studied  Algebra ;  n  times  as  many 
studied  Arithmetic,  and  there  were  m  times  as  many  in  Gram- 
mar as  in  Arithmetic.     How  many  were  in  Algebra  ? 

Ans.  ^L_ 

\-\-n-\-mn 

9.  A  person  said  that  he  was  $450  in  debt.  That  he 
owed  A  a  certain  sum,  B  twice  as  much,  and  C  twice  as  much 
as  to  A  and  B.     How  much  did  he  owe  each  ? 

Ans.  To  A,  $50 ;  to  B,  $100 ;  to  C,  $300. 

10.  A  person  said  that  he  was  owing  to  A  a  certain  sum  ; 
to  B  four  times  as  much  ;  and  to  C  eight  times  as  much  ;  and 
to  D  six  times  as  much  ;  so  that  $570  would  make  him  even 
with  the  world.     What  was  his  debt  to  A  ?  Ans.  $30. 


104  ELEMENTARY   ALGEBRA, 

11.  A  person  said  that  he  was  in  debt  to  four  individuals, 
A,  B,  0,  and  D,  to  the  amount  of  a  dollars  ;  and  that  he  was 
indebted  to  B,  «  times  as  many  dollars  as  to  A ;  to  C,  m  times 
as  many  dollars  as  to  A  ;  and  to  D,  j9  times  as  many  dollars 
as  to  A.     What  was  his  debt  to  A  ? 

Ans.  — dollars. 

l+w-j-m+jo 

12.  If  $75  be  divided  between  two  men  in  the  proportion 
of  3  to  2,  what  will  be  the  respective  shares  ? 

Ans.  $45  and  $30. 

Let x-=-  the  greater  share. 

Then       .     .     .        75 — x=^  the  other. 
To  answer  the  demands  of  the  problem,  we  must  have 
X :  75— ar :  :  3  ;  2 

see  example  13,  (Art.  56).     Observe  the  following  method 

of  solution :  ■ 

Let  3a;=  the  greater  share,  and  2a?  the  smaller  share, 
Then  bx  the  two  shares,  must  equal  the  whole  sum. 
That  is,  5ar=75  or  a:=15.     Therefore,  3.r=45,  the  greater 

share. 

13.  Divide  $150  into  two  parts,  so  that  the  smaller  may 
be  to  the  greater  as  7  to  8.  Ans.  $70;  and  $80. 

14.  Divide  $1235  between  A  and  B,  so  that  A's  share 
may  be  to  B's  as  3  to  2.     Ans,  A's  share,  $741  ;  B's,  $494. 

N.  B.  When  proportional  numbers  are  required,  it  is  gen- 
erally most  convenient  to  represent  them  by  one  unknown 
term,  with  coefficients  of  the  given  relation.  Thus,  numbers 
in  proportion  of  3  to  4,  may  be  expressed  by  2>x  and  Ax,  and 
the  proportion  of  a  to  h  may  be  expressed  by  ax  and  hx. 

15.  Divide  d  dollars  between  A  and  B  so  that  A's  share 
may  be  to  B's  as  m  is  to  »  ^  „^ 

Ans.  As  share, ;    B  s, 

m-\-n  m-\-n 

16.  A  gentleman  is  now  25  years  old,  and  his  youngest 


EQUATIONS.  105 

brother  is  15.     How  many  years  must  elapse  before  tlieir 
ages  will  be  in  the  proportion  of  5  to  4  ?  Ans.  25  years. 

25-l-a;:15+a;::5:4 

1 7.  Two  men  commenced  trade  together ;  the  first  put  in 
S40  more  than  the  second ;  and  the  stock  of  the  first  was  to 
that  of  the  second  as  5  to  4.     What  was  the  stock  of  each  ? 

Ans.  $200;  and  8160. 

18.  A  man  was  hired  for  a  year  for  $100,  and  a  suit  of 
clothes ;  but  at  the  end  of  8  months  he  left,  and  received  his 
clothes  and  $60  in  money,  as  full  compensation  for  the  time 
expired.     What  was  the  value  of  the  suit  of  clothes? 

Ans,  $20. 

19.  Three  men  trading  in  company  gained  $780,  which 
must  be  divided  in  proportion  to  their  stock.  A's  stock  was 
to  B*s  as  2  to  3,  and  A's  to  C's  was  in  the  proportion  of  2  to 
5.     What  part  of  the  gain  must  each  receive  ? 

Ans.  A,  $156  ;  B,  $234;  C,  $390. 

Let 2x=  A's  share  of  the  gain, 

Then 3x=  B's      " 

And 5x=:  C's      <<         »         « 

Therefore,    .     .     .  10a:=780,  or  a:=78. 

20.  A  field  of  864  acres  is  to  be  divided  among  three  farm- 
ers, A,  B,  and  C ;  so  that  A's  part  shall  be  to  B's  as  5  to  11, 
and  C  may  receive  as  much  as  A  and  B  together.  How 
much  must  each  receive  ? 

Ans.  A,  136;  B,  297;  C,  432  acres. 

21.  Three  men  trading  in  company,  put  in  money  in  the 
following  proportion  ;  the  first  3  dollars  as  often  as  the  second 
7,  and  the  third  5.  They  gain  $960.  What  is  each  man's 
share  of  the  gain  ?  Ajis.  $192  ;  $448;  $320. 

22.  A  man  has  two  flocks  of  sheep,  each  containing  the 
same  number ;  from  one  he  sells  80,  from  the  other  20 ; 
then  the  number  remaining  in  the  former  is  to  that  in  the  hit- 


106  ELEMENTARY  ALGEBRA. 

ter  as  2  to  3.     How  many  sheep  did  each  flock  originally 
contain  ?  Ans.  200. 

23.  There  are  two  numbers  in  proportion  of  3  to  4 ;  but 
if  24  be  added  to  each  of  them,  the  two  sums  will  be  in  the 
proportion  of  4  to  5.     What  are  the  numbers  ? 

Ans.  72  and  96. 

24.  A  man's  age  when  he  was  married  was  to  that  of  his 
wife  as  3  to  2  ;  and  when  they  had  lived  together  4  years,  his 
age  was  to  hers  as  7  to  5.  What  were  their  ages  when  they 
were  married?  Ans.  His  age,  24;  hers,  16  years. 

25.  A  certain  sum  of  money  was  put  at  simple  interest, 
and  in  8  months  it  amounted  to  $1488,  and  in  15  months  it 
amounted  to  ^1630.     What  was  the  sum  ?  Ans.  $1440. 

Let  x=  the  sum.  The  sum  or  principle  subtracted  from 
the  amount  will  give  the  interest:  therefore  1488 — x  repre- 
sents the  interest  for  8  months,  and  1530 — x  is  the  interest 
for  15  months. 

Now,  whatever  be  the  rate  per  cent,  double  time  will  give 
double  interest,  &c.     Hence,  8:15::  1488 — x:  1630 — x, 

N.  B.  To  acquire  true  delicacy  in  algebraical  operations,  it 
is  often  expedient  not  to  use  large  numerals,  but  let  them  be 
represented  by  letters.  In  the  present  example,  let  a=1488. 
Then  a-l-42=1530,  and  the  proportion  becomes  8: 15:: a 
— X  :  a4-42 — x. 

Multiply  extremes  and  means,  then 

8a+8'42— 8afel5a~15a7 
HDrop  8a  and  — 8x  from  both  members,  and  we  have 
8«42=7a— 7a; 
Dividing  by  7,  and  transposing,  we  find 

a;=a— 48=1440,  Ans. 

26.  A  certain  sum  of  money  was  put  at  simple  interest  for 
2|-  years,  and  in  that  time  it  amounted  to  $3526,  and  in  38 
months  it  amounted  to  $3606.  What  was  the  sum  put  at 
interest?  Ans.  $3226. 


EQUATIONS.  ,  107 

(Art.  58.)  The  object  of  solving  problems  should  be  to 
acquire  a  knowledge  of  the  utility  and  the  power  of  the 
science,  and  this  knowledge  cannot  be  attained  to  the  fullest 
extent  by  merely  solving  problems ;  we  must  also  learn  how- 
to  propose  them,  and  to  propose  such  as  are  convenient  and 
proper  for  instruction. 

Problem  25  is  extracted  from  an  English  work  ;  and  let  the 
reader  observe  that  the  two  amounts,  $1488,  8 1530,  and 
$1440,  the  sum  put  at  interest,  are  all  whole  numbers,  no 
fraction  of  a  dollar  in  any  of  them,  which  makes  the  problem 
a  neat  and  convenient  one. 

The  question  now  is,  how  the  proposer  discovered  these 
numbers  ?  Did  he  happen  upon  them  ?  Did  he  find  them 
by  repeated  trials  ?  or  did  he  deduce  them  naturally  and  easily 
from  a  scientific  process  ? 

We  can  best  answer  these  questions  by  showing  how  we 
found  the  numbers  to  form  problem  26. 

Wanting  another  example  of  the  same  kind  as  25,  but  of 
different  data,  I  wrote  on  a  slip  of  paper  thus  : 

A  sum  of  money  was  put  at  interest  for  2^  years,  and  the 
amount  for  that  time  was  a  dollars ;  and  for  38  months  the 
amount  was  a-\-d  dollars.     What  was  the  sum  ? 

The  amount  for  38  months  must  be  greater  than  the 
amount  for  30  months,  therefore  c?  is  a  positive  number. 

Let  X  represent  the  sum  lent.  Then  a — x=  the  interest 
for  30  months,  and  a-{-d — x=  the  interest  for  38  months. 

Hence,      .     .     30  :  38  :  :  a — x  :  a-^d — x.  ,^ 

Product  of  extremes  and  means  gives 

30a -f- 30«?— 30a;= 3  8a— 3  8a; 

Dropping  30a  and  — 30a;  from  both  members,  we  have 
30d=Sa — Sx 

Dividing  by  8,  and  transposing,  gives 

a-     a     _- 


108  ,  ELEMENTARY  ALGEBRA. 

Here  a  stands  alone,  and  any  whole  number  greater  than 

—  can  be  written  in  its  place  ;  and  if  we  take  d  of  such  a 
8  ^ 

value  as  to  render  it  divisible  by  8,  the  fraction  —  will  be  a 

^  8 

whole  number,  and  cause  a?  to  be  a  whole  number  also. 

In  preparing  the  example,  I  took  d  equal  80,  then 

8 

is  in  value  300 ;  and  I  took  «,  hap-hazard  at  $3526  ;  there- 
fore, a-l-^=3606,  and  a;=3226,  the  numbers  given  in  the 
problem.  By  taking  different  values  to  a  and  d,  we  may 
form  as  many  numeral  problems  as  we  please  like  problem 
25  or  26  ;  and  if,  in  every  instance,  we  take  care  to  take  d 
of  such  a  value  as  to  render  it  divisible  by  8,  no  fractions 
will  appear  in  the  problems. 

Again,  observe  the  expression  x=za — The  numera- 
tor of  the  fraction  has  30  for  a  coefficient,  and  that  is  the 
number  of  months  that  the  sum  of  money  was  out  at  interest 
before  the  first  amount  was  rendered ;  and  8,  the  denomina- 
tor, is  the  number  of  months  between  the  times  of  rendering 
the  two  amounts. 

Observing  these  facts,  we  may  solve  another  problem  of 
the  like  kind  without  going  through  the  steps  of  the  process. 
For  example. 

27.  A  certain  sum  of  money  was  put  at  simple  interest,  and 
in  13  months  it  amounted  to  a  dollars,  and  in  20  months  it 
amounted  to  a-j-d  dollars.      What  was  the  sum?  Ans.  x. 

And     ...  .     .     x=^a — —         (1) 

7  ^   ^ 

To  form  a  numerical  problem  from  equation  (1),  such  as 
shall  contain  only  whole  numbers,  and  correspond  to  the 
times  here  mentioned,  we  must  take  c?=7,  or  some  midiiple  oj  7. 

Suppose  we  take  c?=  1 4  ;  then 

a-=a— 26  (2) 


EQUATIONS.  109 

Now  it  is  ray  object  to  form  another  numerical  problem  of 
this  kind,  corresponding  to  the  times  mentioned  in  27,  having 
such  numbers  that  the  answer — the  sum  put  at  interest,  shall 
be  just  100  dollars. 

Take  equation  (2),  and  in  place  of  x  write  100,  transpose 
— 26,  and  we  have  a=126,  the  first  amount;  and  as  c?=14, 
a-\-d,  the  second  amount,  must  be  140. 

Hence,  we  may  write  the  problem  thus : 

28.  A  certain  sum  of  money  was  put  at  interest,  and  in  13 
months  the  amount  due  was  ^126,  and  if  continued  at  inter- 
est for  20  months,  the  amount  due  would  have  been  $140. 
What  was  the  sum  put  at  interest  ?  Ans.  $100. 

In  equation  (1),  the  fraction  —  is  the  interest  on  the  sum 

for  13  months,  because  it  is  the  sum,  which,  if  added  to  the 
principal,  will  give  the  amount. 

Here  in  these  problems,  d  and  a  are  perfectly  arbitrary ;  we 

pay  no  attention  to  the  rate  of  interest ;  and  if  we  take  d  of 

any  great  value,  there  will  be  an  unreasonable  quantity  of 

interest ;  and  if  d  is  taken  very  small  in  relation  to  a,  the  rate 

of  interest  will  be  small ;  but  the  algebraist  can  adjust  the 

13d 
rate  by  putting ,  equal  to  any  given  rate  of  interest ;  but 

in  a  work  like  this,  it  is  not  proper  to  carry  these  investiga- 
tions any  further. 

• 

EQUATIONS   CONTINUED. 

(Art.  59.)  Problems  in  which  fractions  mostly  occur. 

1.  The  number  12  is  |  of  what  number  ? 

Ans.  It  is  I  of  the  number  x» 

To  determine  the  numerical  value  of  Xy  we  solve  the  follow- 

3x 
ing  equation,  —=12.     Hence,  16  is  the  number. 


no  ELEMENTARY   ALGEBRA. 

2.  The  number  a  is  |  of  what  number  ?  Am.  ~. 

3 

3.  The  number  21  is  -^  of  what  number  ?  Ans.  49. 

4.  The  number  21  is  the  ^th  part  of  what  number  ? 

Ans.  ^. 


5.  The  number  a  is  the  _ th  part  of  what  number  ? 


n 

an 


m 
6.  If  you  add  together  }  and  |  of  a  certain  number,  the 
sum  will  be  130.     What  is  the  number?  Ans.  420. 

The  following  solution  is  taken  from  another  book,  and  it 
IS  a  fair  specimen  of  the  manner  of  teaching  Algebra,  both 
in  this  country  and  in  England ;  but  in  this  particular  we 
insist  on  improvement. 

♦  Let x=  the  number, 

Then ^-1-~130 

Multiplying  both  members  by  7  and  6,  or  by  42,  we  have 

6x'\''7x=5460 
lSx=5460 
a:=420 

This  is  but  half  Algebra.  An  algebraist  never  multiplies  num- 
bers together,  except  in  final  results,  or  in  some  rare  cases  where  it 
is  impossible  to  do  otherwise. 

To  avoid  this,  let  numbers  be  represented  by  letters ;  and  in 
place  of  130  in  the  equation,  write  a  to  represent  it,  as  taught  in 
(Art.  63.) 

Then -+--a 

Clearing  of  fractions,  6a;+7a;=42a 
Or 18a;=i42a 


EQUATIONS.  HI 

Now,  as  a  is  divisible  by  13,  and  the  quotient  10,  dividing 
both  members  by  13,  gives  a:=420,  without  the  least  effort 
at  numerical  computation. 

It  is  not,  in  fact,  necessary  to  write  a ;  we  may  retain  the 
number  as  a  factor,  or  what  is  better,  take  its  obvious  factors. 

Thus,     ....     6a;4-7^=42-13«10 

Uniting  and  suppressing  the  factors  common  to  both  mem- 
bers, and a;=420 

We  extract  a  solution  to  the  following  problem : 

7.  A  farmer  wishes  to  mix  116  bushels  of  provender,  con- 
sisting of  rye,  barley,  and  oats,  so  that  it  may  contain  f  as 
much  barley  as  oats,  and  \  as  much  rye  as  barley.  How 
much  of  each  must  there  be  in  the  mixture  ? 

Stating  the  question,  Xz=z  oats  ;  and  — =  barley. 
Then, -i-  of  _  is  —  =  rye. 

Forming  the  equation,  a;_j -j- — =116 

Multiplying  by  1 4,  .     .1 4a;-}- 1  Qx-\-  5a;=  1 624 

Uniting  terms, 29a;=1624 

Dividing  by  29,      .....  x=^6Q     the  Arts. 

If  we  keep  the  factors  separate,  we  have 

29a:=n6'14 
Dividing  by  29,  gives  a;=4*  14=56. 

Here  we  find  a  reason  why  the  farmer  wished  to  mix  116  bush- 
els— not  100,  or  115,  or  117 — it  must  be  some  multiple  of  29  to 
have  the  different  kinds  of  grain  come  out  in  whole  numbers. 
Indeed,  the  numbers  in  all  numeral  problems  are  so  chosen  that 
the  final  coefficient  of  the  unknown  quantity  shall  he  some  factor 
in  the  other  member ;  therefore  it  is  worse  than  useless  to  hide  the 
factors  (as  is  often  done),  by  laborious  multiplication. 


112  ELEMENTARY   ALGEBRA. 

8.  Divide  48  into  two  such  parts,  that  if  the  less  be  divided 
by  4,  and  the  greater  by  6,  the  sum  of  the  Quotients  will  be  9. 

Am.  12  and  36. 

9.  A  clerk  spends  §■  of  his  salary  for  his  board,  and  |  of 
the  remainder  in  clothes,  and  yet  saves  $150  a  year.  What 
is  his  yearly  salary  ?  Ans.  $1350. 

10.  An  estate  is  to  be  divided  among  4  children,  in  the 
following  manner : 

The  first  is  to  have  $200  more  than  i  of  the  whole. 

The  second        is  to  have  $340  more  than  ^  of  the  whole. 
The  third  is  to  have  $300  more  than  }  of  the  whole. 

And  the  fourth  is  to  have  $400  more  than  i  of  the  whole. 
"What  is  the  value  of  the  estate  ?  Am.  $4800. 

•11.  Of  a  detachment  of  soldiers,  |  are  on  actual  duty,  | 
of  them  sick,  i  of  the  remainder  absent  on  leave,  and  the 
rest,  which  is  380,  have  deserted.  What  was  the  number  of 
men  in  the  detachment  ?  Am.  2280  men. 

12.  A  man  has  a  lease  for  99  years,  and  being  asked  how 
much  of  it  was  already  expired,  answered  that  |  of  the  time 
past  was  equal  to  |  of  the  time  to  come.  Required  the  tiipe 
past  and  the  time  to  come. 

Assume  a=99.    Am.  Time  past,  54  years. 

13.  It  is  required  to  divide  the  number  204  into  two  such 
parts,  that  2.  of  the  less  being  taken  from  the  greater,  the 
remainder  will  be  equal  to  ^  of  the  greater  subtracted  from  4 
times  the  less.  Am.  The  numbers  are  154  and  50. 

Put  a=204,  and  resubstitute  in  the  result. 

14.  In  the  composition  of  a  quantity  of  gunpowder 
The  nitre  was  10  pounds  more  than  |  of  the  whole. 
The  sulphur     4^  pounds  less    than  ^  of  the  whole. 
The  charcoal      2  pounds  less    than  ■}  of  the  nitre. 
What  was  the  amount  of  gunpowder  ?      Am.  69  pounds. 


EQUATIONS.  lis 

Let ir=  the  -whole. 

Then     ....     ?^-f  10=  the  nitre, 
3 

- — 4i=  the  sulphur, 

22;      10 

2j+-^ — 2=  the  charcoal. 

By  addition,       — _l-4_ — \ .l^^i—^ 

Multiply  both  members  by  6,  and 

4ar4-a;+y+^+21=6a; 

Drop  bx  from  both  members,  then 

4^  .  60  ,  „^ 

-;^  +  y4-21=:r 

Multiply  by  7,  and  drop  4a;  from  both  members. 

And   ....     60+21  •7=3a; 

Dividing  by  3,         204-7«7=a; 

Or, 69=a; 

15.  Divide  $44  between  three  men.  A,  B,  and  C,  so  that 
the  share  of  A  may  be  |  that  of  B,  and  the  share  of  B,  | 
that  of  C.  Ans.  A,  $9  ;  B,  $15,  C,  $20. 

Will  the  student  find  the  reason  why  the  problem  requires 
the  division  of  the  number  44 ;  why  not  45,  47,  or  any  other 
number,  as  well  as  44  ? 

Let  3a;=A's  share,  5a;=B's,  and  y=C's  share, 

3y 

Then -t=^^ 

4 

20a; 
Or y=—~-=z  C's  share. 

o 

Hence,     .     .  3ar-l-5a;+— -=44 
o 

10 


114  ELEMENTARY  ALGEBRA. 

Clearing  of  fractions,  and  uniting  terms,  we  have 

44a:=44*3,  or  x==3 

If  the  problenx  had  required  the  division  of  any  other 
number  of  dollars,  for  instance,  a  dollars,  the  value  of  z 

would  have  been  —  dollars.     Taking  a  equal  44,  or  any  num- 
ber of  times  44,  gives  whole  numhers  for  the  respective  shares. 

16.  What  number  is  that,  to  which,  if  we  add  its  ^,  i,  and 
1,  the  sum  will  be  50  ?  Ans.  24. 

17.  What  number  is  that,  to  which,  if  we  add  its  i,  ^,  and 

A,  the  sum  will  be  a  .^  .        24a 

ns.  -_-. 

18.  If  A  can  build  a  certain  wall  in  10  days,  and  B  can 
do  the  same  in  14  days,  what  number  of  days  will  be  re- 
quired to  build  the  wall,  if  they  both  work  together  ? 

Ans.  5f  days. 

Let  X  represent  the  days  required.     If  A  can  do  the  work 

in  10  days,  in  one  day  he  will  do  -^^  of  it,  and  in  x  days  he 

X  •  X 

will  do  —  of  the  whole  work.     By  the  same  reasoning,   '— 

is  the  part  of  the  work  done  by  B. 


Therefore,  —-+—-  =  1.     (1  is  the  whole  work). 
10     14  ^  ' 

19.  If  A  can  do  a  piece  of  work  in  a  days,  and  B  can  do 
the^same  in  b  days,  how  long  will  it  take  them,  if  they  both 
work  together?  ^^^    _^  days. 

I  now  wish  to  propose  a  numerical  problem  in  all  respects 
like  problem  18,  except  that  the  number  of  days  shall  be  a 
whole  number,  and  the  answer  shall  be  8. 

The  answer  to  1 9  is  a  r/eneral  answer ;   and  now  if  we 


EQUATIONS.  115 

require  a  particular  answer,  8,  we  simply  require  the  verifi- 
cation of  the  following  equation. 

Or ah=Sa-]-Sh 

In  this  equation,  if  we  assume  a,  the  equation  will  give  b, 
or  if  we  assume  b,  the  equation  will  give  a  corresponding 
value  to  a.     But  whichever  letter  we  assume,  it  must  be 
assumed  greater  than  8  ;  because  it  requires  either  man  more 
than  8  days  to  do  the  work,  for  they  together  do  it  in  8  days. 
Now,  assume  a=12,  then  the  equation  becomes 
125=8'12+85 
46=8'12 
5=24 

We  can  now  write  out  our  numerical  problem  thus  : 

20.  A  can  do  a  piece  of  work  in  1 2  days ;  B  can  do  the 

same  in  24  days.     How  many  days  will  be  required,  if  they 

both  work  together  ?  Ans.  8. 

2t.  A  young  man,  who  had  just  received  a  fortune,  spent 

f  of  it  the  first  year,  and  ^  of  the  remainder  the  next  year ; 

when  he  had  $1420  left.     What  was  his  fortune  ? 

Ans.  811360. 

22.  If  from  ^  of  my  hight  in  inches,  12  be  subtracted,  i 
of  the  remainder  will  be  2.    What  is  my  hight  ? 

Ans.  5  feet  6  inches. 

23.  A  laborer.  A,  can  perform  a  piece  of  work  in  5  days, 
B  can  do  the  same  in  6  days,  and  C  in  8  days ;  in  what  time 
can  the  three  together  perform  the  same  work  ? 

Ans.  2/^  days. 
Let  a;=  the  number  of  days  in  which  all  three  can  do  it. 

24.  After  paying  out  i  and  }  of  my  money,  I  had  remain- 
ing 66  guineas.     How  many  guineas  had  I  at  first  ? 

Ans.  120. 


116  ELEMENTARY  ALGEBRA. 

25.  In  a  certain  orchard,  i  are  apple  trees,  |  peach  trees, 
J-  plum  trees,  100  cherry  trees,  100  pear  trees.  How  many 
trees  in  the  orchard  ?  Ans.  2400. 

26.  A  farmer  has  his  sheep  in  jfive  different  fields,  viz  :  ^ 
in  the  first  field,  ^  in  the  second,  }  in  the  third,  ^j  in  the 
fourth,  and  45  in  the  fifth  field.     How  many  sheep  in  the 

Jock?  Ans.  120. 

27.  A  person  at  play,  lost  ^  of  his  money,  and  then  won  3 
shilhngs ;  after  which  he  lost  i  of  what  he  then  had  ;  and,  on 
counting,  found  that  he  had  12  shillings  remaining.  What 
had  he  at  first?  Ans.  20  shilhngs. 

(Art.  60).  When  equations  contain  compound  fractions,  and 
simple  ones,  clear  them  of  the  simple  fractions  first,  and  unite, 
as  far  as  possible,  all  the  simple  terms. 

We  give  a  few  examples  to  show  the  advantage  of  observ- 
ing this  expedient. 

1.  Given ; — = to  find  the  value  of  a;. 

9  6x-\-3  3 

Multiply  all  the  terms  by  the  smallest  denominator,  3. 
That  is,  divide  all  the  denominators  by  3,  and 

3     ^2a:-{-l  ^ 

Multiplying  again  by  3,  and  dropping  6a;-{-7  from  both 
members,  we  have 

21a;— 39_ 

~2x-{-l"~ 
Clearing  of  fractions,  transposing,  (fee,  we  find  a;=4. 

^    ^.         7a:+16       ar-f  8    ,a: 

2.  Given-^-=^^-^j+-tofind.. 

Multiply  by  21,  and  from  both  members  drop  7x,  then 

^      21a:-|-21'8 
16= ! 

4a^— 11 

Clearing  of  fractions 

And     .     .     .      64a:— llM6=21a;+21'8 


EQUATIONS.  II? 

For  the  purpose  of  showing  something  of  the  spirit  of 
Algebra,  we  will  put  a=8:  after  dropping  21a;  from  both 
members, 

Then     .     .      43a;— 11 -2^=2  la 
.    Or   .     .     .  43a;— 22a=21a 

Or    .     .     .     .  43a;=43a    or    ar=a=8. 

„    ^.         9a;4-20     4a; — 12  ,  x 

3.  Given  — ^-~  =— +-,  to  find  x, 

36  5x — 4      4 

Multiplying  by  36,  and  dropping  9a;  from  both  members. 

Then 20=??(i^l-l) 

5x — 4 

4.  Given  -\ p^  =^ -^^ — ■ ^  to  find  x. 

X        ax-f-ox  a-\-b 

Multiply  by  a;. 

Then    .     .  a-6+?^=?=(5^?^(.5!f±!^ 
a-\-b  a-J-5 

Multiply  by  (a-\-b),  and  unite  known  quantities, 

Then     ...       3a^—2b^=(3a^—2b^)(3a'+2b^)x 

1 

^~3a^-f2b^' 

5x-\-5         9       6x — 12 

5.  Given  ■ — r-^-r    1   = —   to  find  a;.     .  Ans.  a?=2. 

x-\-2         4         X — 2 

^     -,.         2a;+l      402— 3a;  471— 6a;  ^     .    , 

6.  C^iven  —^- ig""^"^ 2 ^^  ^^^  ^• 

-4ws.  a;=72. 
^    _.         18a;— 19  ,  lla;-|-21     9a;+15  ^     .    , 

Ans.  a;=7. 


^    ^.         20a;+36  ,  5a;+20     4a;  ,  „  ■       , 

8.  Given  _^_+^_=_+3xx  tofind^.. 


u4w5.  x=^4. 


«    ^.             6         3a;  ,  2     2a;-}-5     3  ,    , 

9.  Given  -^---+-=~3 to  find  a;. 


Ans.  x=^. 


118  ELEMENTARY  ALGEBRA. 

I  Sar-f- 1 8 

10.  Given \-5-{-x=z — - —   to  find  a:.     .  Ans.  x=6. 

X — 5  3 

11.  Divide  the  number  48  into  two  such  parts,  that  7 
divided  by  one  part  shall  be  equal  to  5  divided  by  the  other 
part.     Required  the  parts.  Ans.  28  and  20. 

1 2.  Divide  the  number  48  into  two  such  parts,  that  one  may 
be  to  the  other  as  7  to  5.     Required  the  parts.  Ans.  28  and  20. 

13.  A  person  in  play,  lost  a  fourth  of  his  money,  and  then 
won  back  3  shillings  ;  after  which  he  lost  a  third  of  what  he 
now  had,  and  then  won  back  2  shillings ;  lastly,  he  lost  a 
seventh  of  what  he  then  had,  and  then  found  he  had  but  1 2 
shillings  remaining.     What  had  he  at  first  ? 

Ans.  20  shillings. 

14.  A  shepherd  was  met  by  a  band  of  robbers,  who  plun- 
dered him  of  half  of  his  flock  and  half  a  sheep  over.  After- 
ward a  second  party  met  him,  and  took  half  of  what  he  had 
left,  and  half  a  sheep  over  ;  and  soon  after  this,  a  third  party 
met  him  and  treated  him  in  like  manner  ;  and  then  he  had  5 
sheep  left  ?     How  many  sheep  had  he  at  first  ? 

Ans.  47  sheep. 

15.  A  man  bought  a  horse  and  chaise  for  341  (a)  dollars. 
Now,  if  f  of  the  price  of  the  horse  be  subtracted  from  twice 
the  price  of  the  chaise,  the  remainder  will  be  the  same  as  if 
4  of  the  price  of  the  chaise  be  subtracted  from  3  times  the 
price  of  the  horse.     Required  the  price  of  each. 

Ans.  Horse,  $152;  chaise,  $189. 
N.  B.  Let  8a;=  the  price  of  the  horse. 
Or  let     .    7a:=  the  price  of  the  chaise.  ^ 

Solve  this  question  by  both  of  these  notations. 

16.  A  laborer  engaged  to  serve  for  60  days^  on  these  con- 
ditions :  That  for  every  day  he  worked  he  should  have  75 
cents  and  his  board,  and  for  every  day  he  was  idle  he  should 
forfeit  26  cents  for  damage  and  board.  At  the  end  of  the 
time  a  settlement  was  made,  and  he  received  $25.  How 
many  days  did  he  work,  and  how  many  days  was  he  idle  ? 


EQUATIONS.  119 

The  common  way  of  solving  sucli  questions  is  to  let  a'= 
the  days  he  worked ;  then  60 — x  represents  the  days  he  was 
idle.     Then  sum  up  the  account  and  put  it  equal  to  $25. 

Another  method  is,  to  consider  that  if  he  worked  the  whole 
60  days,  at  75  cents  per  day,  he  must  receive  ^45.  But  for 
every  day  he  was  idle,  he  not  only  lost  his  wages,  75  cents, 
but  25  cents  in  addition.  That  is,  he  lost  %\  every  day  he 
was  idle. 

Now,  let  rr=  the  days  he  was  idle.  Then,  x=-  the  dollars 
he  lost.     And  45 — a:=25  or  a;=20,  the  days  he  was  idle. 

17.  A  person  engaged  to  work  a  days  on  these  conditions: 

For  each  day  he  worked  he  was  to  receive  h  cents  ;  for  each 

day  he  was  idle  he  was  to  forfeit  c  cents.     At  the  end  of 

a  days  he  received  d  cents.     How  many  days  was  he  idle  ? 

.        ah — d  . 
Ans.  -rr-i —  days. 

Let  x=  the  number  of  days  he  was  idle. 

Had  he  worked  every  day  he  must  have  received  ah  cents. 
But  for  every  idle  day  we  must  diminish  this  sum  by  (6+c) 
centsr;  and  for  x  days,  the  diminution  must  be  (b-\-c)x  cents. 

That  is,  al) — (b-\-c)x=d  by  the  question. 

,T  cib — d 

Hence, x=:—-, — 

18.  A  boy  engaged  to  convey  30  glass  vessels  to  a  certain 
place,  on  condition  of  receiving  5  cents  for  every  one  he  de- 
livered safe,  and  forfeiting  12  cents  for  every  one  he  broke. 
On  settlement,  he  received  99  cents.  How  many  did  he 
break?  Ans.  3. 

19.  A  boy  engaged  to  carry  n  glass  vessels  to  a  certain 
place,  on  condition  of  receiving  a  cents  for  every  one  he  deliv- 
ered, and  to  forfeit  b  cents  for  every  one  he  broke.  On  set- 
tlement he  received  d  cents.     How  many  did  he  break  ? 

Ans.  The  number  represented  by  —ri" 


120  ELEMENTARY  ALGEBRA. 

SIMPLE    EQUATIONS 

CONTAINING    TWO    UNKNOWN    QUANTITIES. 

(Art.  61.)  We  have  thus  far  considered  such  equations 
only  as  contained  but  one  unknown  quantity ;  but  we  now 
suppose  the  pupil  sufficiently  advanced  to  comprehend  equa- 
tions containing  two  or  more  unknown  quantities. 

There  are  many  simple  problems  which  one  may  meet  with 
in  Algebra,  which  cannot  be  solved  by  the  use  of  a  single 
unTcnown  quantity,  and  there  are  also  some  which  may  he 
solved  by  a  single  letter,  that  may  become  much  more  simple 
by  using  two  or  more  unknown  quantities. 

When  two  unknown  quantities  are  used,  two  independent 
equations  must  exist,  in  which  the  value  of  the  unknown  let- 
ters must  be  the  same  in  each.  When  three  unknown  quan- 
tities are  used,  there  must  exist  three  independent  equations, 
in  which  the  value  of  any  one  of  the  unknown  letters  is  the 
same  in  each. 

In  short,  there  must  he  as  mxiny  independent  equations  as 
unknown  quantities  used  in  the  question. 

An  independent  equation  may  be  called  a  primitive  or  prime 
equation — one  that  is  not  derived  from  any  other  equation. 
Thus,  x-\-2>y=a,  and  2a;+6y=2a,  are  not  independent  equa- 
tions, because  one  can  be  derived  from  the  other ;  but  x-\-2)y 
=a,  and  Ax-\-by=h,  are  independent  equations,  because 
neither  one  can  be  reduced  to  the  other  by  any  arithmetical 
operation. 

The  reason  that  two  equations  are  required  to  determine 
two  unknown  quantities,  will  be  made  clear  by  considering 
the  following  equation : 

a;+y=20 

This  equation  will  be  verified  if  we  make  x—\,  and  y=19, 
ora:=2,  and  2/=  18,  or  *=^,  and  y=19i,  &c.,  &c.,  without 
limit.     But  if  we   combine  another  equation  with  this,  as 


EQUATIONS.  121 

gc — ^y=4,  then  we  have  to  verify  two  equations  with  the  same 
values  to  x  and  y,  and  only  one  value  for  x  and  one  value  for 
y  will  answer  both  conditions. 

Thus,  .....     a;4-y=20 

And X — y—  4 

By  addition    .     .     .         2a:=24 

x=\2y  and  y=8. 

That  is,  we  have  found  a  value  for  x  and  another  to  y  (12 
and  8),  so  that  their  sum  shall  be  20,  and  their  difference  4 ; 
and  no  other  possible  numbers  will  answer. 

A  merchant  sends  me  a  bill  of  16  dollars  for  3  pairs  of  shoes 
and  2  pairs  of  hoots ;  afterward  he  sends  another  bill  of  23 
dollars  for  4  pairs  of  shoes  and  3  pairs  of  boots,  charging  at  the 
same  rate.  What  was  his  price  for  a  pair  of  shoes,  and  what 
for  a  pair  of  boots  ? 

This  can  be  resolved  by  one  unknown  quantity,  but  it  is  far 
more  simple  to  use  two. 

Let    x=  the  price  of  a  pair  of  shoes. 

And  y=  the  price  of  a  pair  of  boots. 

Then  by  the  question  3a;+2z/=16 

And     .....    4a:+3y=23 

These  two  equations  are  independent ;  that  is,  one  cannot 
be  converted  into  the  other  by  multiphcation  or  division,  not- 
withstanding the  value  of  x  and  of  y  are  the  same  in  both 
equations. 

Equations  are  independent  when  they  express  dififerent  con- 
ditions, and  dependent  when  they  express  the  same  conditions 
under  different  forms. 

To  reduce  equations  involving  two  unknown  quantities,  it 
is  necessary  to  perform  some  arithmetical  operation  upon 
them,  which  will  cause  one  of  the  unknown  quantities  to  dis- 
appear.    These  operations  are  called  ehmination. 

There  are  three  principal  methods  of  eliminaHcn. 
11 


122  ELEMENTARY  ALGEBRA. 

1.  By  comparison.  2.  By  substitution.  3.  By  addition  or 
^subtraction. 

All  the  operations  rest  on  the  axioms. 

FIRST   METHOD. 

(Art.  62.)  Transpose  the  terms  containing  y  to  the  right 
hand  sides  of  the  equations,  and  divide  by  the  coefficients  of 
X,  and 

From  equation  {A)  we  have  x= (  C) 

o 

And  from  {B)  we  have     .     a:= (B) 

Put  the  two  expressions  for  x  equal  to  each  other  (Ax  7), 

,     ,  16— 2y     23— 3y 

And -= 

3  4 

An  equation  which  readily  gives  ^=-5,  which,  taken  as  the 
value  of  y  in  either  equation  (  0)  or  (B),  will  give  x=2. 


SECOND    METHOD. 

(Art.  63.)  To  explain  the  second  method  of  elimination, 
resume  the  equations 

3x+2y=16^       (A) 
4a;-l-3y=23         (B) 

The  value  of  x  from  equation  (A)  is  ic=i(16 — 2y). 

Substitute  this  value  for  x  in  equation  (B),  and  we  have 
4X1(16 — 2y)-\-3y=23,  an  equation  containing  only  y. 

Reducmg  it,  we  find  y=5,  the  same  as  before. 

Observe,  that  this  method  consists  in  finding  the  value  of 
one  of  the  unknown  quantities  from  one  equation,  and  substi- 
tuting that  value  in  the  other.  Hence,  it  is  properly  called 
the  method  by  substitution. 


^      EQUATIONS.  1'2S 

THIRD   METHOD    OF    ELIMxNATION. 
(Art.  64.)  Resume  again   3x-\-2y  =  16         (A) 
4a;+3y=23         (JB) 

When  the  coefficients  of  either  a;  or  y  are  the  same  in  both 
equations,  and  the  signs  alike,  that  term  will  disappear  by 
subtraction. 

When  the  signs  are  unlike,  and  the  coefficients  equal,  the 
term  will  disappear  by  addition. 

To  make  the  coefficients  of  x  equal,  multiply  each  equation  hy 
the  coefficient  of  x  in  tJte  other. 

To  make  the  coefficients  of  y  equal,  multiply  each  equation  by 
i/ie  coefficient  of  y  in  the  other. 

Multiply  equation  (A)  by  4  and  12ii;+8y=64 
Multiply  equation  (B)  by  3  and  12a?+9y=69 

Difference y=5,  as  before. 

To  continue  this  investigation,  let  us  take  the  equations 
2ar-f3y=23         (A) 
5x—2y=10         (B) 

Multiply  equation  (A)  by  2,  and  equation  (B)  by  3,  and 
we  have  4x-\-6y=46 

15x—6y—30 

Equations  in  which  the  coefficients  of  y  are  equal,  and  the 
signs  unlike.  In  this  case  add,  and  the  y's  will  destroy  each 
other,  giving 19a;=76 

Or x=4 

Of  these  three  methods  of  ehmination,  sometimes  one  is 
preferable  and  sometimes  another,  according  to  the  relation 
of  the  coefficients  and  the  positions  in  which  they  stand. 

No  one  should  be  prejudiced  against  either  method ;  and  in 
practice  we  use  either  one,  or  modifications  of  them,  as  the 


124  ELEMENTARY  ALGEBRA. 

case  may  require.     The  forms  may  be  disregarded  when  the 
principles  are  kept  in  view. 

EXAMPLES. 

1.  Given  ]  *"i/=^^^  (^U  to  find,  and  y. 

From  (A)      ....     x=-^^ 
From  (JB)      ....     a;=16— 4y 
Therefore      .     .       — ^^=16— 4y     (Ax.  7). 

Ans.  y=2,  x=S. 
(   7x-{-4y=5S  (A)  ) 

2-  Given  -j  9^_4^_33  ^^^   [  to  find  a:  and  y. 

Ans.  x=6,  y=4. 

Here,  it  would  be  very  inexpedient  to  take  the  first  method 
of  elimination. 

Observe  that  the  coefficients  of  y  are  alike  in  number,  but 
opposite  in  signs. 

A  skillful  operator  takes  great  advantage  of  circumstances, 
and  very  rarely  goes  through  all  the  operations  of  set  rules  ; 
but  this  skill  can  only  be  acquired  by  observation  and 
practice. 

Add  the  two  equations.     Why  ? 

(  5x+6y=5S  (A)   )        ,    , 

3.  Given  j  ^^_^q^^^^  ^^'^   \  to  find  x  and  y. 

Ans.  ir=8 ;  y=3. 
Subtract  ( J5)  from  (A).    Why  ? 

4.  Given  j       Tll'^^\o    iiU^^-^---^y- 

Ans.  x=4;  y=2. 
Add  (^^and  (i?).    Why? 

5.  Given  j  f  i^^=^f  1  p|  [  to  find  .  and  y. 

(  3a:+4y=88  (B)  f 

Ans.  x=S ;  y=16. 
Multiply  (B)  by  2.    Why  ? 


EQUATIONS.  125 

C.  Given  i   „    ;  ..^  r  to  find  x  and  z, 

Ans.      x=7 ;  z=8. 

^     _.         C  4x-\-  6y=  46  ) 

7.  Given  S    ^         ^         .  ^  r  to  nnd  a;  and  y. 

(  5a:—  2?/=   10  j  ^ 

„    ^.  (   2x+  3y=  31    )        ^   ^ 

8.  Given  1    ^         „         .  „  f  to  find  a?  and  y. 

(  4a; —  3y=   17  ) 

Ans.      x=S ;  y=5. 

(  4y-|-     2?=102  )        ,    , 

9.  Given  i         ,     ,         ,^   h  to  find  y  and  z. 

(     3/4-  42=  48   )  ^ 

Ans.     ?/=24 ;  2=6 

«    I  .^       ^^   r  to  find  a;  and  y. 
8a;+10y=26  )  ^ 

^715.  a:=2 :  y=l. 

(    6y+  3a?=93  )   ,    ^   ^ 
11-  Given]  ^^_^^^._g^^tofindyanda.. 

Ans.  y=\2;  a?=ll. 

12.  Given  Aar-|-^y==14,  and  Ja:-1-Jy=ll,  to  find  x  and  y, 

^725.  a;=24;  3^=6. 

13.  Given  x-\-y=S,  and  lx-\-y=7,  to  find  a;  and  y. 

Ans.  x=s^6;  y=4. 

14.  Given  ja;+7y=99,  and  ^y-f  7a;=51,  to  find  x  and  y. 

Ans.  x=7  ;  y  —  14. 

PEOBLEMS  PRODUCING   EQUATIONS   OF   TWO   UNKNOWN   QUANXITIES. 

1.  A  man  bought  3  bushels  of  wheat  and  5  bushels  of  rye 
for  38  shillings ;  and  at  another  time,  6  bushels  of  wheat  and 
3  bushels  of  rye  for  48  shillings.  What  was  the  price  for  a 
bushel  of  each  ? 

Let  x=  price  of  wheat,  and  y=  price  of  rye. 

By  the  first  condition,  3a;-l-5y=38  (A) 

By  the  second,     .     .   6a:+3y=48  {J3) 

Ans.  x=6 ;  y='U 


126  ELEMENTARY   ALGEBRA. 

2.  A  gentleman  paid  for  6  pairs  of  hoots,  and  4  pairs  of 

shoes,  $44  ;  and  afterward,  for  3  pairs  of  boots,  and  7  pairs 
of  shoes,  $32.     What  was  the  price  of  each  per  pair  ? 

Ans.  Boots,  $6  ;  shoes,  $2. 

3.  A  man  spends  30  cents  for  apples  and  pears,  buying  his 
apples  at  the  rate  of  4  for  a  cent,  and  his  pears  at  the  rate  of 
5  for  a  cent.  He  afterward  let  his  friend  have  half  of  his 
apples  and  one-third  of  his  pears,  for  13  cents,  at  the  same 
rate.     How  many  did  he  buy  of  each  sort  ? 

Let  x=  the  number  of  apples, 
y=    «         «       of  pears, 
A  cent  =  the  price  of  1  apple ;  hence,  x  apples  are  worth 


-  cents.     Therefore, 
4 

"By  the  first  condition,  .     . 

^  l=- 

(^) 

By  the  second,        .     .     . 

i+f5=- 

(-8) 

Multiplying  (B)  by  2, 

I+S=- 

(1) 

Subtracting  (1)  from  (A), 

5     15 

(2) 

Multiplying  (2)  by  15,    . 

Sy— 22/=60, 

or   y=60 

This  value  of  y  put  in  (^),  gives  --f  12=30,  or  ar=72. 

4.  What  fraction  is  that,  to  the  numerator  of  which,  if  1  be 
added,  its  value  will  ^,  but  if  1  be  added  to  the  denominator, 
its  value  will  be  ^  ? 

Let cc=  the  numerator. 

And y==  the  denominator, 

X 

Then —  ==•  the  fraction, 

y 


EQUATIONS,  127 

And  we  shall  have  the  two  equations. 

Clearing  of  fractions,  3a;+3=y  (1) 

And 4a;— l=y  (2) 

Taking  (1)  from  (2),    ic— 4=0,    or    x=4. 

Hence,  from  1  we  find         y=15,  and  the  fraction  is  j-\. 

5.  What  fraction  is  that,  to  the  numerator  of  which,  if  4  he 
added,  the  value  is  i,  but  if  7  be  added  to  its  denominator, 
its  value  will  be  |  ?  Ans.  j\. 

G.  A  and  B  have  certain  sums  of  money :  says  A  to  B, 
**  Give  me  ^15  of  your  money,  and  I  shall  have  five  times  as 
much  as  you  have  left."  Says  B  to  A,  **  Give  me  ^5  of 
your  money,  and  I  shall  have  exactly  as  much  as  you  have 
left."     How  many  dollars  had  each  ? 

Ans.  A  had  ^35 ;  and  B,  ^25. 

7.  What  fraction  is  that  whose  numerator  being  doubled, 
and  its  denominator  increased  by  7,  the  value  becomes  |; 
but  the  denominator  being  doubled,  and  the  numerator 
increased  by  2,  the  value  becomes  -j  ?  Ans.  a. 

8.  If  A  give  B  ^5  of  his  money,  B  will  have  twice  as  much 
money  as  A  has  left ;  and  if  B  give  A  ^5,  A  will  have  thrice 
as  much  as  B  has  left.     How  much  had  each  ? 

Ans.  A  had  $13;  B,  $11. 

9.  A  merchant  has  sugar  at  9  cents  and  at  13  cents  a 
pound,  and  he  wishes  to  make  a  mixture  of  100  pounds  tliat 
shall  be  worth  12  cents  a  poimd.  How  many  pounds  of  each 
quality  must  he  take  ?  ' 

Ans.  25  pounds  at  9  cents,  and  75  pounds  at  13  cents. 

10.  A  person  has  a  saddle  worth  ;i650,  and  two  horses. 
When  he  saddles  the  poorest  horse,  the  horse  and  saddle  are 
worth  twice  as  much  as  the  best  horse  ;  but  when  he  saddles 


128  ELEMENTARY  ALGEBRA. 

the  best,  the  horse  and  saddle  are  together  worth  three  times 
the  other.     What  is  the  value  of  each  horse  ? 

Ans.  Best,  MO ;  poorest,  ^30. 

1 1 .  One  day  a  gentleman  employs  4  men  and  8  boys  to 
labor  for  him,  and  pays  them  40  shillings ;  the  next  day  he 
hires  at  the  same  rate,  7  men  and  6  boys,  for  50  shillings. 
What  are  the  daily  wages  of  each  ? 

Ans.  Man's,  5  shillings ;  boy's,  2  shillings  6  pence. 

1 2.  A  merchant  sold  a  yard  of  broadcloth  and  3  yards  of 
velvet,  for  $25;  and,  at  another  time,  4  yards  of  broadcloth 
and  5  yards  of  velvet,  for  ^65.  What  was  the  price  of  each 
per  yard?  Ans.  Broadcloth,  $10;  velvet,  $5. 

13.  Find  two  numbers,  such  that  half  the  first,  with  a  third 
part  of  the  second,  make  9,  and  a  fourth  part  of  the  first,  with 
a  fifth  part  of  the  second,  make  5.  Ans.  8  and  15. 

14.  A  gentleman  being  asked  the  age  of  his  two  sons,  an- 
swered, that  if  to  the  sum  of  their  ages  18  be  added,  the 
result  will  be  double  the  age  of  the  elder ;  but  if  6  be  taken 
from  the  difference  of  their  ages,  the  remainder  will  be  equal 
to  the  age  of  the  younger.     What  were  their  ages  ? 

Ans.  30  and  12. 

15.  A  says  to  B,  **  Give  me  100  of  your  dollars,  and  I 
shall  have  as  much  as  you."  B  replies,  **Give  me  100  of 
your  dollars  and  I  shall  have  twice  as  much  as  you.  How 
many  dollars  has  each  ?  ^725.  A,  $500;  B,  $700. 

16.  Find  two  numbers,  such  that  f  of  the  first  and  f 
of  the  second  added  together,  will  make  1 2  ;  and  if  the  first 
be  divided  by  2,  and  the  second  multiplied  by  3,  |  of  the  sum 
of  these  results  will  be  26.  Ans.  15  and  10^. 

17.  Says  A  to  B,  "^  of  the  difference  of  our  money  is 
equal  to  yours  ;  and  if  you  give  me  $2,  I  shall  have  five 
times  as  much  as  you."     How  much  has  each  ? 

Ans.  A,  $48;  B,  $12. 

18.  A  market-woman  bought  eggs,  some  at  the  rate  of  2 
for  a  cent,  and  some  at  the  rate  of  3  for  2  cents,  to  the  amount 


EQUATIONS.  129 

of  65  cents.  She  afterward  sold  them  all  for  100  cents, 
thereby  gaining  half  a  cent  on  each  egg.  How  many  of  each 
kind  did  she  buy  ?  Ans.  50  of  one  ;  60  of  the  other. 

19.  What  two  numbers  are  those,  whose  sum  is  a  and  dif- 
ference b  ? 


Let  x==  the  greater. 

Ans.  The  greater  is 

a  b 
2+2 

y=  the  less. 

The  less  is 

a  b 
£-2 

Sum  is     .     . 

a 

Difference  is 

h. 

(Art.  64.)  From  the  result  of  this  problem,  we  learn  one 
important  fact,  which  will  be  of  use  to  us  in  solving  other 
problems. 

The  fact  is  this :  That  the  half  sum  of  any  two  nmnbers, 
added  to  the  half  difference,  is  the  greater  of  the  two  numbers  ; 
and  the  half  sum,  diminished  by  the  half  difference ^  gives  the 
less. 

20.  There  are  two  numbers  whose  sum  is  100,  and  three 
times  the  less  taken  from  twice  the  greater,  gives  150  for 
remainder.     What  are  the  numbers  ?  Ans.  90  and  10. 

The  half  sum  of  the  two  numbers  is  50.  Now  let  xz;=  the 
half  difference.       Then  50-|-  ar=  the  greater  number. 

And 50 —  a;=  the  less  number. 

Twice  the  greater  is  \00-\-2x 
Three  times  less  is  1 50 — 3.r 
Difference  is      .      — 50-}-52;=150. 

21.  What  two  numbers  are  those  whose  sum  is  12,  and 
whose  product  is  35  ?  A71S.  7  and  5. 

Let Q-\-x  =  the  greater, 

Then 6 — x  =  the  less. 


Product,     ....  36— a''=35.     Hence,  x=^\. 


130  ELEMENTARY  ALGEBRA. 

22.  What  two  numbers  are  those  whose  difference  is  4,  and 
product  96  ?  Ans.   12  and  8. 

Let  x=  the  half  sum  ;  2=  the  half  difference. 

23.  The  difference  of  two  numbers  is  6,  and  the  sum  of 
their  squares  is  50.     What  are  the  numbers  ?    Ans.  7  and  1. 

24.  The  difference  of  two  numbers  is  8,  and  their  product 
is  240.     What  are  the  numbers  ?  A7is.  12  and  20. 

(Art.  65.)  To  reduce  equations  containing  three  or  more 
unknoAvn  quantities,  we  employ  the  same  principles  as  for  two 
unknown  quantities,  and  no  more  ;  and  from  these  principles 
we  draw  the  following 

Rule  . —  Combine  any  one  of  the  equations  vnih  each  of  the 
others,  so  as  to  eliminate  the  same  unknown  quantity ;  there  will 
thus  arise  a  new  set  of  equations  containing  one  less  unknown 
quantity. 

In  the  same  manner  combine  one  of  these  new  equations  with 
each  of  the  others,  and  thus  obtain  another  set  of  equations  con- 
taining one  less  unknown  quantity  than  the  last  set ;  and  so 
continue,  until  an  equation  is  found  containing  one  unknown 
quantity ;  solving  this  equation,  and  substituting  the  value  of  iis 
unknovm  quantity  in  the  other  equations,  the  other  unknown 
quantities  are  easily  found. 

EXAMPLES. 

(  x-\-  y-f-  ^=  9  "^ 
1.  Given   <  a;+2y4- 3^=16  >   to  find  x,  y,  and  z. 
(  x-\-3y+4z=21  ) 

By  the  first  method,  transpose  the  terms  containing  y  and 
z  in  each  equation,  and 

x=  9 —  y —  z 
x=zl6—2y—3z 
x=2\—Sy—4z 


EQUATIONS.  iM 

Then  putting  the  1st  and  2d  values  equal,  and  the  2d  and 
3d  values  equal,  gives 

9 —  y—  ^=16 — 2y— 32 
1 6—2?/ — 3z = 2 1  — 3y — 4z 
Transposing  and  condensing  terms, 

And y=7 — 2z 

A-lso, y=5 —  z 

Hence,      ....    5 — z=7 — 2z,  or  z=2. 

Having  z=2,  we  have  y=5 — 2=3,  and  having  the  values 
of  both  z  and  y,  by  the  first  equation  we  find  x=4.  - 

r  2a?+4y— 32=22  ^ 

2.  Given  I  4x — 2?/4-52=18  >  to  find  x,  y  and  z. 

(  6ir-f7y—  2=63  ) 
Multiplying  the  first  equation  by  2,     .     4a;+8y —  62=44 
And  subtracting  the  second,       .     .     .     4x — 2y-f-  52=18 

The  result  is,     (.4) lOy— -112=26 

Then  multiply  the  first  equation  by  3,    6x-\-12y —  92=66 

And  subtract  the  third, 6x-\-  'ly —    2=63 

The  resxilt  is,     (^) 5y—  82=  3 

Multiply  the  new  equation,  {B),  by  2,  lOy — 162=  6 

And  subtract  this  from  equation  (A)y  \0y — 1 12=26 

The  result  is, 52=20 

Therefore,  2=4. 

Substituting  the  value  of  z  in  equation  {B),  and  we  find 
y=7. 

(    3:r+9y+82=41  \ 

3.  Given  \    5x-\-Ay — 22=20  \  to  find  x,  y,  and  2. 

(  ILT+Ty— 62=37  ) 

Ans.  x=2;  y=S ;  2=1. 

(Art.  66.)  When  three,  four,  or  more  unknown  quantities, 
w^ith  as  many  equations,  are  given,  and  their  coefficients  are 


132 


ELEMENTARY   ALGEBRA. 


all  prime  to  each  other,  the  operation  is  necessarily  long. 
But  when  several  of  the  coefficients  are  multiples,  or  mea- 
sures of  each  other,  or  unity,  several  expedients  may  be  re- 
sorted to  for  the  purpose  of  facilitating  calculation. 

No  specific  rules  can  be  given  for  mere  expedients.  Exam- 
ples alone  can  illustrate.  Some  few  expedients  will  be  illus- 
trated by  the  following 


EXAMPLES. 


■x-\-y-{-z=zm-\ 
1.  Given  •{  x-\-y — 2=25  )■  to  find  x,  y,  and  z. 


X — y — z=  9  J 
Subtract  the  2d  from  the  1st,  and  22=6. 
Subtract  the  3d  from  the  2d,   and  2?/=  16. 
Add  the  1st  and  3d,  and     .     .       2;r=40. 

2.  Given  \  x — y      =  4  V  to  find  ar,  y,  and  z. 

yx — z       =  6J 

Add  all  three,  and  3:r=36,  or  a:=12. 

r  x—y—z=  6-^ 

3.  Given  \  3y — x — z—  1 2  V  to  find  x,  y,  and  z.  .  . 

I  'lz—y—x=^^] 

4.  Given   a;4-iy=100,    y-\-\z^\m,   2+^0;=  100,  to  find 
X,  y,  and  z. 

Puta=100.  .     Ans.  a:=64;  y=72  ;  and  2=84. 

'  u-\-v-\-x-\-y=\0 

u-\-v-\-z-\-x=:^  1 1 
6.  Given  ■{  u-\-v-\-z-\-x=12  S-  to  find  the  value  of  each. 

u-\-x-{-y-\-z—13 

v-]-x-\-y-\-z='i4^ 

Here  are  Jive  letters  and  five  equations.     Each  letter  has 


EQUATIONS.  133 

the  same  coefficient,  one  understood.     Each  equation  has  4 
letters,  z  is  wanting  in  the  1st  equation,  y  in  the  2d,  &c. 

Now  assume  u-]rv-\-x-]ry-]rz=^s. 

Then 5—2=10        {A) 

s—y=\\ 

s—x—l^ 

5— v=13 

s — w=14 
Add,  and     ....   5s—s=Q0     Or  «=  15. 

Put  this  value  of  «  in  equation  {A),  and  2=5,  &c. 

6.  Given  ir+y=a,  a;-l-2=6,  2^4-0=c. 

Add  the  1st  and  2d,  and  from  the  sum  subtract  the  3d. 
Am.  a;=i(a+5 — c),  y=l{a-\-c — 5),  z—\(J)-{-c — a). 


5.  Reduce  the 
equations 


6.  Reduce  the 
equations 


'  x-\-y=5^ 
y+  2=82 

2-1- 2^=68 

u-\-x=32 


ix-^3y=23 

.+    ^=8 

J/+ 32=31 
x-\-y  +2-l-2it'=39 


Ans. 


7.  Reduce  the 
equations 

8,  Reduce  the 
equations 


r4ar-|-2y— 32=  4 
}  3a;— 5y+22=22 
I   x-{-  y-{-  2=12 


•i.r+iy-f  2=46  >| 

\x—  y-hzi=  9  I 

a^+|y—i2=19J 


-4w«. 


Ans. 


Ans, 


a:=20, 
y=32, 
2=50, 
m;=18, 
w=12. 

a;=6, 

I    2=8, 

lw=9. 


ar=5, 

y-1, 

2=6. 

a:=20, 
y=12, 
2=32. 


134  ELEMENTARY  ALGEBRA 

QUESTIONS    PRODUCING    EQUATIONS 

CONTAINING    THREE    OR    MORE    UNKNOWN    QUANTITIES 

1.  There  are  three  persons,  A,  B,  and  C,  whose  ages  are  as 
follows :  If  B's  age  be  subtracted  from  A's,  the  difference 
will  be  C's  age  ;  if  five  times  B's  age  and  twice  C's,  be  added 
together,  and  from  their  sum  A's  age  be  subtracted,  the 
remainder  will  be  147.  The  sum  of  all  their  ages  is  96. 
What  are  their  ages?  Ans.  A's,  48;  B's,  33;  C's,  15. 

2.  Find  what  each  of  three  persons.  A,  B,  and  C,  is  worth, 
from  knowing,  1st,  that  what  A  is  worth  added  to  3  times 
what  B  and  C  are  worth  make  4700  dollars;  2d,  that  what 
B  is  worth  added  to  4  times  what  A  and  C  are  worth  make 
5800  dollars ;  3d,  that  what  C  is  worth  added  to  5  times 
what  A  and  B  aie  worth  make  6300  dollars. 

Ans.  A  is  worth  8500;  B,  8600;  C,  8800. 

3.  A  gentleman  left  a  sum  of  money  to  be  divided  among 
four  servants,  so  that  the  share  of  the  first  was  ^  the  sura 
of  the  shares  of  the  other  three ;  the  share  of  the  second,  J- 
of  the  sum  of  the  other  three ;  and  the  share  of  the  third,  J- 
the  sum  of  the  other  three  ;  and  it  was  found  that  the  share 
of  the  last  was  14  dollars  less  than  that  of  the  first.  What 
was  the  amount  of  money  divided,  and  the  shares  of  each 
respectively  ? 

Ans.  The  sum  was  8120 ;  the  shares,  40,  30,  24  and  26. 

4.  The  sum  of  three  numbers  is  59;  |  the  difference  of 
the  first  and  second  is  5,  and  ^  the  difference  of  the  first  and 
third  is  9;  required  the  numbers.  Ans.  29,  19,  and  11. 

5.  There  is  a  certain  number  consisting  of  two  places,  a 
unit  and  a  ten,  which  is  four  times  the  sum  of  its  digits,  and 
if  27  be  added  to  it,  the  digits  will  be  inverted.  What  is  the 
number?  Ans.  36. 

Note. — Undoubtedly  the  reader  has  learned  in  Arithme- 
tic that  numerals  have  a  specific  and  a  local  value,  and  every 


EQUATIONS.  135 

remove  from  the  unit  multiplies  by  10.  Hence,  if  x  repre- 
sents a  digit  in  the  place  of  tens,  and  y  in  the  place  of  units, 
the  number  must  be  expressed  by  10x-\-y.  A  number  con- 
sisting of  three  places,  with  x,  y,  and  z  to  represent  the  digits, 
must  be  expressed  by  100a;-l-]0?/+s. 

6.  A  number  is  expressed  by  three  figures ;  the  sum  of 
these  figures  is  1 1  ;  the  figure  in  the  place  of  units  is  double 
tliat  in  the  place  of  hundreds,  and  when  297  is  added  to  this 
number,  the  sum  obtained  is  expressed  by  the  figures  of  this 
number  reversed.     What  is  the  number  ?  Ans.  326. 

7.  Divide  the  number  90  into  three  parts,  so  that  twice 
the  first  part  increased  by  40,  three  times  the  second  part 
increased  by  20,  and  four  times  the  third  part  increased  by 
10,  may  be  all  equal  to  one  another. 

A71S.  First  part,  35;  second,  30;  and  third,  25. 

If  the  object  is  merely  to  solve  the  seventh  example,  it 
would  not  be  expedient  to  use  three  unknown  symbols. 
Let  ^(x — 40)=  the  first  part,  &c. 

8.  Find  three  numbers,  such  that  the  first  with  J-  of  the 
other  two,  the  second  with  1  of  the  other  two,  and  the  third 
with  I  of  the  other  two,  shall  be  equal  to  25. 

Ans.   13,  17,  and  19. 

9.  A  man  with  his  wife  and  son,  talking  of  their  ages,  said 
that  his  age,  added  to  that  of  his  son,  was  16  years  more 
than  that  of  his  wife ;  the  wife  said  that  her  age,  added  to 
that  of  her  son,  made  8  years  more  than  that  of  her  husband ; 
and  that  all  their  ages  added  together  amounted  to  88  years. 
What  was  the  age  of  each  ? 

Ans.  Husband,  40,  wife,  36,  and  son  12  years. 

10.  There  are  three  numbers,  such  that  the  first,  with  i 
the  second,  is  equal  to  14;  the  second,  with  i  part  of  the 
third,  is  equal  to  18;  and  the  third,  with  i  part  of  the  first,  is 
equal  to  20;  required  the  numbers.  Ans.  8,  12,  and  18. 


136  ELEMENTARY  ALGEBRA. 

11.  Find  three  members,  such  that  i  of  the  first,  i  of  the 
second,  and  }  of  the  third  shall  be  equal  to  62;  i  of  the  first, 
I  of  the  second,  and  i  of  the  third  equal  to  47;  and  i  of  the 
first,  i  of  the  second,  and  ^  of  the  third  equal  to  38. 

Ans.  24,  60,  and  120. 

12.  There  are  two  numbers,  such  that  i  the  greater  added 
to  1  the  lesser,  is  13;  and  if  i  the  lesser  is  taken  from  ^  the 
greater,  the  remainder  is  nothing.     Required  the  numbers. 

Ans.  18  and  12. 

12.  Find  three  numbers  of  such  magnitude,  that  the  first 
with  the  ^  sum  of  the  other  two,  the  second  with  ^  of  the 
other  two,  and  the  third  with  i  of  the  other  two,  may  be  the 
same,  and  amount  to  51  in  each  case.     Ans.  16,  33,  and  39. 

14.  A  said  to  B  and  C,  **  Give  me,  each  of  you,  4  of  your 
sheep,  and  I  shall  have  4  more  than  you  will  have  left."  B 
said  to  A  and  C,  **If  each  of  you  will  give  me  4  of  your 
sheep,  I  shall  have  twice  as  many  as  you  will  have  left."  C 
then  said  to  A  and  B,  "Each  of  you  give  me  4  of  your 
sheep,  and  I  shall  have  three  times  as  many  as  you  will  have 
left."     How  many  had  each  ?  Ans.  A,  6;  B,  8;  C,  10. 

16.  A  person  bought  three  silver  cups ;  the  price  of  the  first, 
with  ^  the  price  of  the  other  two,  was  26  dollars ;  the  price 
of  the  second,  with  i  of  the  price  of  the  other  two,  was  26 
dollars ;  and  the  price  of  the  third,  with  i  the  price  of  th*» 
other  two,  was  29  dollars ;  required  the  price  of  each. 

Ajis.  $8,  $18,  and  $16 

16.  A's  age  is  double  that  of  B's,  and  B's  is  triple  that  of 
C's,  and  the  sum  of  all  their  ages  is  140;  what  is  the  age  of 
each?  Ans.  A's  =84;  B's  =42;  C's  =14. 

17.  A  man  wrought  10  days  for  his  neighbor,  his  wife  4 
days,  and  son  3  days,  and  received  1 1  dollars  and  60  cents ; 
at  another  time  he  served  9  days,  his  wife  8  days,  and  his  son 
6  days,  at  the  same  rates  as  before,  and  received  1 2  dollars  ; 
a  third  time  he  served  7  days,  his  wife  6  days,  his  son  4  days, 


EQUATIONS.  137 

at  the  same  rates  as  before,  and  received  9  dollars.     What 
were  the  daily  wages  of  each  ? 

A71S.  Husband's  wages,  81.00;  wife,  0;  son,  50  cts. 

(Art.  67.)  In  this  last  example  we  put  x  to  represent  the 
daily  wages  of  the  husband,  y  the  wages  of  the  wife,  and  z 
the  wages  of  the  son,  and  in  conclusion,  y  was  found  equal  to 
0 ;  but  it  might  have  came  out  a  minus  quantity,  and  if  it  had, 
it  would  have  shown  that  the  presence  of  the  wife  was  not  a 
source  of  income,  but  expense ;  and  if  correct  results  are 
given  at  the  settlements,  the  si^ns  to  the  different  quantities 
will  show  whether  any  particular  individual  received  wages 
or  was  on  expense  for  board,  as  in  the  following  problems  : 

18.  A  man  worked  for  a  person  ten  days,  having  his  wife 
with  him  8  days,  and  his  son  6  days,  and  he  received  10  dol- 
lars and  30  cents  as  compensation  for  all  three ;  at  another 
time  he  wrought  12  days,  his  wife  10  days,  and  son  4  days, 
and  he  received  13  dollars  and  20  cents  ;  at  another  time  he 
wrought  15  days,  his  wife  10  days,  and  his  son  12  days,  at 
the  same  rates  as  before,  and  he  received  4 3  dollars  85  cents. 
What  were  the  daily  wages  of  each  ? 

Ans.  The  husband  75  cts.;  wife,  50  cts.  The  son  20  cts. 
expense  per  day. 

Here  the  language  of  the  problem  is  improper,  as  It  im- 
plies that  all  received  wages ;  but  the  solution  shows  that  this 
could  not  be  the  case  ;  for  the  value  of  the  son's  wages  comes 
out  minus,  which  is  opposition  to  plus  or  to  positive  wages,  that 
is,  expense. 

A  stronger  illustration  of  this  principle  will  be  shown  by 
the  following  problem : 

19.  Two  men,  A  and  B,  commenced  trade  at  the  same 
time  ;  A  had  3  times  as  much  money  as  B,  and  continuing  in 
trade,  A  gains  400  dollars,  and  B  150  dollars;  now  A  has 
twice  as  much  money  as  B.  How  much  did  each  have  at 
first? 

12 


138  ELEMENTARY  ALGEBRA. 

Without  any  special  consideration  of  the  question,  it  imphes 
that  both  had  money,  and  asks  how  much.  But  on  resolving 
the  question  with  x  to  represent  A's  money,  and  y  B's,  we 
find x= — 300 

And 2/=— 100  dollars. 

That  is,  they  had  no  money,  and  the  minus  sign  in  this 
case  indicates  debt;  and  the  solution  not  only  reveals  the  nu- 
merical values,  but  the  true  conditions  of  the  problem,  and 
points  out  the  necessary  corrections  of  language  to  correspond 
to  an  arithmetical  sense. 

That  is,  the  problem  should  have  been  written  thus : 

A  is  three  times  as  much  in  debt  as  B ;  hut  A.  gains  400  dol- 
lars, and  B  150  ;  now  A  has  twice  as  much  money  as  B.  How 
much  were  each  in  debt  ? 

As  this  enunciation  corresponds  with  the  real  circumstances 
of  the  case,  we  can  resolve  the  problem  without  a  minus  sign 
in  the  result.     Thus  : 

Let     x=  B's  debt,  then       3ar=  A's  debt. 
150 — x=  B's  money,  400 — 2>xz=  A's  money. 
Per  question,  400— 3a; =300— 2a:.     Or  a;=]00. 

20.  What  number  is  that  whose  fourth  part  exceeds  its 
third  part  by  12  ?  Ans.  —144. 

But  there  is  no  such  abstract  number  as  — 144,  and  we 
cannot  interpret  this  as  debt.  It  points  out  error  or  impos- 
sibUity,  and  by  returning  to  the  question  we  perceive  that  a 
fourth  part  of  any  number  whatever  cannot  exceed  its  third 
part ;  it  must  be,  its  third  part  exceeds  its  fourth  part  by  1 2, 
and  this  enunciation  gives  the  positive  number,  144.  Thus 
do  equations  rectify  subordinate  errors,  and  point  out  special 
conditions. 

21.  A  man  when  he  was  married  was  30  years  old,  and  his 


EQUATIONS.  139 

wife  15.     How  many  years  nmst  elapse  before  his  age  will  be 
three  times  the  age  of  his  wife  ? 

Ans.  The  question  is  incorrectly  enmiciated ;  7|-  years 
before  the  marriage,  not  after,  their  ages  bore  the  specified 
relation. 

22.  What  fraction  is  that  which  becomes  |  when  1  is 
added  to  its  numerator,  and  becomes  f  when  1  is  added  to 
its  denominator  ? 

A71S.  In  an  arithmetical  sense,  there  is  no  such  fraction. 
The  algebraic  expression,  Zxj>  "^iH  give  the  required  results. 

23.  Divide  the  number  10  into  two  such  parts  that  their 
product  shall  be  50. 

Let ^4-y=  the  greater  number. 

And x — y=  the  less. 

Then 22^=10,    or    x=5. 

The  product  of  the  two  numbers  is  x^ — y^,  and  by  the 
question  must  be  equal  to  50. 

That  is,    ....      x^—y^—50. 

But,  x=5;  hence,  a^=25,  which,  drop  from  both  members. 

And —f=25 

Or y*=— 25 

That  is,  the  question  calls  for  two  equal  factors  whose  pro- 
duct is  minus  25;  but  equal  factors  will  never  give  a  minus 
product;  there  is  no  2d  root  of — 25,  and  the  value  of  the 
unknown  quantity  in  such  cases  is  said  to  be  imaginary, 
which  shows  that  the  problem  is  impossible. 

Here,  y=±5^ — 1,  a  value  that  has  no  existence  in 
numbers. 


SECTION    III. 


INTOLUTION. 

(Art.  68.)  Equations,  and  the  resolution  of  problems  pro- 
ducing equations,  do  not  always  result  in  the  first  powers  of 
the  unknown  (Quantities,  but  different  powers  are  frequently  in- 
volved, and  therefore  it  is  necessary  to  investigate  methods  of 
resolving  equations  containing  higher  powers  than  the  first ; 
and  preparatory  to  this  we  must  learn  involution  and  evolution 
of  algebraic  quantities. 

(Art.  69.)  Involution  is  the  method  of  raising  any  quan- 
tity to  a  given  power.  Evolution  is  the  reverse  of  involution, 
and  is  the  method  of  determining  what  quantity  raised  to  a 
proposed  power  will  produce  a  given  quantity. 

As  in  Arithmetic,  involution  is  performed  by  multiplication, 
and  evolution  by  the  extraction  of  roots. 

The  first  power  is  the  root  or  quantity  itself. 

The  second  power,  commonly  called  the  square,  is  the 
quantity  multiplied  by  itself. 

The  third  power  is  the  product  of  the  second  power  by  the 
quantity. 

The  fourth  power  is  the  third  power  multiphed  into  the 
quantity,  &c. 

Or  we  may  consider  the  second  power  of  a  quantity  to  be 
the  quantity  taken  twice  as  a  factor. 

The  third  power  is  the  quantity  taken  three  tirties  as  a  factor. 


INVOLUTION.  141 

The  fourth  power  is  tlie  quantity  taken  four  times  as  a 
factor. 

The  tenth  power  is  the  quantity  taken  ten  times  as  a  factor ; 
and  so  on  for  any  other  power. 

The  nth.  would  be  the  quantity  taken  n  times  as  a  factor. 

Thus,  let  a  represent  any  quantity. 

Its     frst     power  is  .     .     .     a=a 


Its     second  power  is  . 

a*a=c^ 

The  third    power  is  . 

.  a'a'a=a^ 

The  fourth  power  is  . 

a*a*a*a=^a^ 

The  fifth     power  is   a*a*a'a*a=^c^ 

In  general  terms  a  to  the  n\h  power  is  a'a,  &c.,  =a",  and 
n  may  be  any  number  whatever. 

(Art.  70.)  When  the  quantity  is  negative,  all  the  odd 
powers  will  be  minus,  and  all  the  even  powers  will  be  plus. 

For,  by  the  rules  of  multiplication, 

— aX— a=+a* 

And     .     .     .     — aX — aX — a= — a^ 

&c.=<&c. 

(Art.  71.)  When  we  require  the  5th  power  of  a,  we 
simply  write  a^;  when  the  8th  power  we  write  a^,  &c.,  for 
any  other  power. 

That  is,  a  is  the  same  as  a\  the  exponent  1  is.  understood  ; 
and  when  we  require  the  nth.  power  of  a,  we  conceive  its 
exponent  1  understood  multiplied  by  n,  which  makes  a". 

The  second  power  of  a?  is   a^Xc^=a^. 

The  third    power  of  a?  is  a^'a^*a^=a^. 

The  tenth    power  of  a^  is  a^^. 

The  nth      power  of  a?  is  a^". 

That  is,  it  is  the  exponent  of  the  quantity  repeated  as  many 
times  as  there  are  units  in  the  index  of  the  power. 

Thus,  the  7th  power  of  d^  has  the  exponent  of  a  (4)  re* 
peated  7  times,  and  the  result  is  a*. 


142 


ELEMENTARY  ALGEBRA, 


From  thig  we  derive  the  following  rule  to  raise  a  single 
quantity  to  any  power. 

Rule  . — Midtiply  the  exponent  of  iJie  qxtaniity  hy  the  index 
qf  the  required  power. 


EXAMPLES 

Raise  a^      to  the    3(1  power.  . 

Raise  i/*      to  the  4th  power.  . 

Raise  i"    to  the  5th  power.  . 

Raise  a?     to  the  4th  power.  . 

Raise  y'      to  the    3d  power.  . 

Raise  x^     to  the  6th  power.  . 

Raise  x^     to  the  mth  power.  . 

Raise  aa?    to  the    3d  power.  . 

Raise  ah^x'^  to  the    2d  power.  . 

Raise  cV   to  the  5th  power.  . 


.  Ans.  2^ 
.  Am.  y^. 
.  Ans,  I^\ 
.  Ans.  x^^ 
.  Ans.  y^V 
.  Ans.  a^^ 
,  Ans.  a;""^ 
.  Ans.  aV 
Ans.  d^b'^3?. 
Ans.   c^y 


(Art.  72.)  By  the  definition  of  powers,  the  second  power 
is  any  quantity  multiplied  by  itself;  hence  the  second  power 
of  ax  is  a^x^,  the  second  power  of  the  coefficient  a,  as  well  as 
the  other  quantity  x;  but  a  may  be  a  numeral,  as  6x,  and  its 
second  pcywer  is  36x^.  Hence,  to  raise  any  simj^le  quantity  to 
any  power,  raise  the  numeral  coefficient,  as  in  Arithinetic,  to 
Vie  required  power,  and  annex  the  powers  of  the  given  literal 
quantities. 


EXAMPLES. 

1.  Required  the   Sd  power  of  3ax^. 

2.  Required  the  4th  power  of  \y^. 

3.  Required  the   3d  power  of  — 2x. 

4.  Required  the  4th  power  of  — 3a: 
6.  Required  the  2d  power  of  8a^i'. 
6.  Required  the   3d  power  of  b:^z. 


Ans.  27a^x'^ 
Ans.  If/ 
Ans.  — 8ar* 
Ans.  Blx'^ 
Am.  Gia'^b^. 
Ans.  125x«^. 


INVOLUTION.  143 

7.  Required  the   3d  power  of  Qcc'^f-x.     .     Ans.  2\^a}^tf^. 

8.  Required  the  4tli  power  of  2am^     .     Ans.     16a^6V. 

(Art^  73.)  When  the  quantity  to  be  raised  to  a  power  is 
a  fi'action,  we  must  observe  the  rules  for  the  multiplication 
of  fractions,  and  multiply  numerators  by  numerators,  and 
denominators  by  denominators.* 

Thus,  the  2d  power  of  -  is  — — -=  — 

Hence,  to  raise  fractions  to  powers,  we  have  the  following 
Rule. — Raise  both  numerator  and  denominator  to  the  re- 
quired power. 

EXAMPLES. 

Observe,  that  by  the  rules  laid  down  for  multiplication,  the 
even  powers  of  minus  quantities  must  be  plus,  and  the  odd 
powers  minus. 

9.a%^  4^4^12 

1.  Required  the   2d  power  of -— ~.    .     .     .     Ans.  ~p-j* 

2.  Required  the  6th  power  of  — --.    .     .     .     Ans.  — — -• 


*  Suppose  we  were  required  to  raise  r  to  the  fifth  power,  and  did  not 

know  whether  the  denominator  was  to  be  raised  or  not,  we  could  decide 
the  point  by  means  of  an  equation,  as  follows  : 

The  fraction  has  soine  value  which  we  represent  by  a  symbol,  say  P. 
Then  P=r'     Now  if  we  can  find  the  true  5th  power  of  P,  it  will  be 

the  required  5th  power  of  the  fraction. 

Clearing  the  equation  of  fractions,  we  have 

bP=a 
Taking  the  5th  power  of  both  members  gives 

a> 
By  division,       .     .     .     .     .     P^—j-. 

b" 

This  equation  shows  that  to  raise  any  fraction  to  any  power,  the 

numerator  and  denominator  must  be  raised  to  that  power 


14 

ELEMENTARY  ALGEBRA. 

3. 

Required  the  6tli  power  of  —  •       •     • 

Am.  -   ^   -. 

4. 

Required  the  6\h.  power  of  |a^5.     .     . 

Am.  4>« 

5. 

Required  the   2d  power  of  -g.  .     .     . 

.           9 
.     .  Ans.      -.' 
a* 

6. 

Required  the   3d  power  of  -^. 

.     .  Ans.  -r--' 

7. 

2y 
Required  the  4th  power  of  — —.    .     . 

16/ 

8. 

Required  the   3d  power  of      ^.    •     • 

(Art.  74.)  The  powers  of  compound  quantities  are  raised 
by  the  application  of  the  rule  for  compound  multiplication, 
(Art.  14). 

Let  a-^b  be  raised  to  the  2d,  3d,  4th,  &c.,  powers, 
a +5 
a-\-b 

a  +a6 

ab-^b'' 


2d  power  or  square,    a^-^-^ab  ~\-b^ 
a-\-b 


a?-^S>a^b-\-ab^ 

a^b-\-^aJ)'-\-b^ 


3d  power  or  cube,       a?-{-2>d^b-{-?>aI^-\-b^ 
a-\-b 


a^+2>a%-\-2,a^b''-\-ab'' 

a?b-\-Sa%^-\-^ab''-\-b'' 


The  4th  power,      .     a^-{-^a^b^Q(rb'-^^ab^-\-h' 
a-\-b 


a^J[.^a%-\-Qa:'b''+^d'b^-\-ab'' 

a%-{-Aa?b^-\-Qa^b^-{-Aad'-^b'' 


The  6th  power,      .     a^+5a:^b-{-\0a?b''-\'\0a^b^-\-bab'^-\-b^ 


INVOLUTION.  145 

By  inspecting  the  result  of  each  product,  we  may  arrive  at 
general  principles,  according  to  which  any  power  of  a  bino- 
mial may  be  expressed,  without  the  labor  of  actual  multipli- 
cation. This  theorem  for  abbreviating  powers,  and  its  general 
application  to  both  powers  and  roots,  first  shown  by  Sir  Isaac 
Newton,  has  given  it  the  name  of  Newton's  binomial,  or  the 
binomial  theorem. 

Observations. — Observe  the  5th  power :  a,  being  the  first, 
is  called  the  leading  term ;  and  h,  the  second,  is  called  the 
following  term.  The  sum  of  the  exponents  of  the  two  let- 
ters in  each  and  all  of  the  terms  amount  to  the  index  of 
the  power.  In  the  5th  power,  the  sum  of  the  exponents  of  a 
and  b  is  5;  in  the  4th  power  it  is  4;  in  the  10th  power  it 
would  be  10,  &c.  In  the  2d  power  there  are  three  terms  ;  in 
the  3d  power  there  are  4  terms ;  in  the  4th  power  there  are 
5  terms  ;  always  one  more  term  than  the  index  of  the  power 
denotes. 

The  2d  letter  does  not  appear  in  the  first  term;  the  1st 
letter  does  not  appear  in  the  last  term. 

The  highest  power  of  the  leading  term  is  the  index  of  the 
given  power,  and  the  powers  of  that  letter  decrease  by  one 
from  term  to  term.  The  second  letter  appears  in  the  2d  term, 
and  its  exponent  increases  by  one  from  term  to  term,  aS  the 
exponent  of  the  other  letter  decreases. 

The  8th  power  of  (a-\-b)  is  indicated  thus,  (a-{-6)^     Wlien 
expanded,  its  literal  part  (according  to  the  preceding  obser- 
vations) must  commence  with  a^,  and  the  sum  of  the  expo- 
nents of  every  term  amount  to  8,  and  they  will  stand  thus 
a\  a%  a^h\  a%\  a'b\  a'b\  d'b\  ab\  b^ 

The  coefiicients  are  not  so  obvious.  However,  we  observe 
that  the  coefficients  of  the  first  and  last  terms  must  be  uniiy 
The  coefficients  of  the  terms  next  to  the  first  and  last  are 
equal,  and  are  the  same  as  the  index  of  the  power.  The 
coefficients  increase  to  the  middle  of  the  series,  and  then 
decrease  in  the  same  manner,  and  it  is  manifested  that  there 
13 


146 


ELEMENTARY   ALGEBRA. 


must  be  some  law  of  connection  between  the  exponents  and 
the  coefficients. 

By  inspecting  the  6th  power  of  a-{-b,  we  find  that  the  2d 
coefiicient  is  5,  and  the  3d  is  10. 


5X4 


10 


The  3d  coefiicient  is  the  2d,  multiplied  by  the  exponent  of 
the  leading  letter,  and  divided  hy  the  exponent  of  the  second 
letter  increased  hy  unity. 

In  the  same  manner,  the  fourth  coefficient  is  the  third,  mul- 
tiplied by  the  exponent  of  the  leading  letter,  and  divided  by 
the  exponent  of  the  second  letter  increased  by  unity,  and  so 
on  from  coefiicient  to  coefficient. 


The  4th  coefficient  is 

The  6th  is  .     .     .     . 

The  last  is  .     .     .     . 
Now  let  us  expand    . 
For  the  1st  term  write 
For  the  2d  term  write 

8X7 
2 

28X6 


10X3_ 

3 
10X2_ 

4 
5X1 

10 


5 


1  understood. 


For  the  3d, 
For  the  4th, 
For  the  6th, 


=28 


3 
56X6 


^d'b 
28a«52 

lOa'b* 


Now,  as  the  exponents  of  a  and  b  are  equal,  we  have 
arrived  at  the  middle  of  the  series,  and  of  course  to  the  high- 
est coefficient.  The  coefficients  now  decrease  in  the  reversa 
order  in  which  they  increased. 


INVOLUTION.  147 

Hence,  the  expanded  power  is 

Let  the  reader  observe,  that  the  exponent  of  b,  mcreased 
by  unity,  is  always  equal  to  the  number  of  terms  from  the 
beginning  or  from  the  left  of  the  power.  Thus,  5^  is  in  the 
3d  term,  &c.  Therefore  in  finding  the  coefficients,  we  may 
divide  by  the  number  of  terms  already  written,  in  place  of 
the  exponents  of  the  second  term  increased  by  unity. 

If  the  binomial  {a-\-b)  becomes  (a-j-l),  that  is,  when  b 
becomes  unity,  the  8th  power  becomes, 

a«+8a^-|-28a«+56a°+70a*+56a3H-28a24-8a+l. 
•    Any  power  of  1  is  1,  and  1  as  a  factor  never  appears. 

If  a  becomes  1,  then  the  expanded  power  becomes, 
H-8S+2862+56&'-l-706^+5655+285«-f-8i^+i«. 

The  manner  of  arriving  at  these  results  is  to  represent  the 
unit  by  a  letter,  and  expand  tlie  simple  literal  terms,  and  after- 
ward substitute  their  values  in  the  result. 

(Art.  74.)  If  we  expand  (a — b)  in  place  of  (a-|-i),  the 
exponents  and  coefficients  will  be  precisely  the  same,  but  the 
principles  of  multiplication  of  quantities  afifected  by  different 
signs  will  give  the  minus  sign  to  the  second  and  to  levery 
alternate  term. 

Thus,  the  6th  power  of  (a — b)  is 

a^—Qdb-[- 1  ba%^—ma%^-\- 1 5a'b'—6al^+b\ 

(Art.  75.)  This  method  of  readily  expanding  the  powers 
of  a  binomial  quantity  is  one  apphcation  of  the  *'  binomial  theo- 
rem,'' and  it  was  thus  by  induction  and  by  observations  on 
the  result  of  particular  cases  that  the  theorem  was  established. 
Its  rigid  demonstration  is  somewhat  difficult,  but  its  applica- 
tion is  simple  and  useful. 

Its  most  general  form  may  arise  from  expandmg  {a-\-'b)^. 

When  w=3,  we  can  readily  expand  it. 

When  71=4,  we  can  expand  it. 


148  ELEMENTARY  ALGEBRA. 

Wlien  71=  any  whole  positive  number,  we  can  expand  it. 

Kow  let  us  operate  with  n  just  as  we  would  with  a  known 
number,  and  we  shall  have 

ji \ 

1 

We  know  not  where  the  series  would  terminate,  until  we 

know  the  value  of  n.     We  are  convinced  of  the  truth  of  the 

result,  when  n  represents  any  positive  whole  number  ;  but  let 

n  be  negative  or  fractional,  and  we  are  not  so  sure  of  the 

result. 

The  result  would  be  true,  however,  whatever  be  the  value 
of  n;  but  this  requires  demonstration,  and  a  deeper  investi- 
gation than  it  would  be  proper  to  go  into  in  a  work  like  this. 

When  w  is  a  fraction,  the  operation  is  extracting  a  root  in 
place  of  expanding  a  power. 

But  for  the  demonstration  of  the  binomial  theorem,  and  its 
application  to  the  extraction  of  roots,  we  refer  the  reader  to 
our  University  Edition  of  Algebra. 

EXAMPLES. 

1.  Expand  {x-^yf Ans.  ar'+Sar'y-f Sa^-fy*. 

2.  Expand  (y-f-^)^. 

3.  Required  the  third  power  of  3a;-|-2y. 

We  cannot  well  expand  this  by  the  binomial  theorem, 
because  the  terms  are  not  simple  literal  quantities.  But  we 
can  assume  Sx=a  and  2y=b.     Then 

3a:+2y«a+i,  and  (a-t-i)3=a'+3«'i+3a6'-f  *" 


INVOLUTION.  149 

Kow  to  return  to  the  values  of  a  and  h,  we  have, 

3ab^=:3 XSxX  4y^=36xy^ 


Hence,    (3x+2yy=27u^+54x'2j-\-3Gxy'-\-Qf. 

4.  Required  the  4th  power  of  2a^ — 3. 

Let  x=:2a^,  y=o.     Then  expand  (x — yy,  and  return  the 
values  of  x  and  y,  and  we  shall  find  the  result. 
16a«— 96a«+216a^— 216a2+81. 

5.  Required  the  cube  of  (a-^b-\-c-\-d). 

As  we  can  operate  in  this  summary  manner  only  on  bino- 
mial quantities,  we  represent  a-\-b  by  x,  or  assume  x=a-]-b, 
and  y=c-\-d. 

Then  {xi-yy=a^^Sx'y+3xy^-\-f' 

Returning  the  values  of  x  and  y,  we  have 
(a+by+3(a-]-by{c+d)+3{a+bXc-{-dy+(c-\-dy. 

Now  we  can  expand  by  the  binomial,  these  quantities  con- 
tained in  parentheses. 

6.  Required  the  4th  power  of  2a-\-3x. 

Ans.  16a''-\-96a^x+216a^x^-{-216ax^-\-81x\ 

7.  Expand  (x'^-{-37fy. 

Ans.  a;^°+ 1 5a;«y2_|_9o^6y^270;2;y4.405a;y+2432/i°. 

8.  Expand  {^a'+axy.          Ans.  8a«+12a^a;+6aV+aV. 

9.  Expand  (x—iy. 

Ans.  x^—6x^-\-15x'*—20x''-\-15x^—6x-\-l. 

10.  Expand  (3a:— 5)^         Ans.  27x^—135x^-{-225x^—n5. 

11.  Expand  (2a— 55)=^.      Ans.  Sa^—60a''b-{-150ab^—125bK 

12.  Expand  {4a'b—2c^y. 

Ans.  25Ga}^b*—512a%h^+3Ma^Pc'^—128(^bc^- 1  &<?, 


ISO  ELEMENTARY  ALGEBKA. 


EVOLUTION. 

(Art.  76.)  Evolution  is  tlie  converse  of  involution. 

Involution  is  the  expanding  of  roots  to  powers.  Evolutidte 
is  extracting  the  root  when  the  power  is  given. 

To  find  rules  for  operation,  we  must  observe  how  powers 
are  formed,  and  then  we  shall  be  able  to  trace  the  alterations 
hack.  Thus,  to  square  a,  we  double  its  exponent,  which 
makes  a^,  (Art.  71).  The  square  of  a^  is  a"*,  the  cube  of  a^ 
is  a°,  &c.  Take  the  4th  power  of  x,  and  we  have  x'^.  The 
wth  power  of  ^^  is  x^^,  &c.,  &c. 

Now,  if  mivltiplying  exponents  raises  simple  literal  quanti- 
ties to  powers,  dividing  exponents  must  extract  roots.     Thus, 

2. 

the  square  root  of  a**  is  a^.     The  cube  root  of  d^  must  be  a^. 
The  cube  root  of  a  must  have  its  exponent,  (1  understood) 

divided  by  3,  which  will  make  a^. 

Therefore,  roots  are  properly  expressed  hy  frojctional  exponents. 

.  The  square  root  of  a  is  d^,  and  the  exponents,  J ,  i,  i,  &c., 
indicate   the   third,   fourth,   and   fifth  roots.     The  6th  root 

of  or'  is  x^  ;  hence,  we  perceive  that  the  numerators  of  tht 
exponent  indicate  the  power  of  the  quantity,  and  the  denom- 
nator  the  root  of  that  power. 

(Art.  77.)  The  square  of  ax  is  a^x^.  We  square  both 
factors,  and  so,  for  any  other  powers,  we  raise  all  the  factors 
to  the  required  power.  Conversely,  then,  we  extract  roots 
by  taking  the  required  roots  of  all  the  factors.  Thus  the 
cube  root  of  8a;^  is  2x. 

The  square  root  of  64a'*  is  obviously  8a^  and  from  these 
examples  we  draw  the  following  rule  for  the  extraction  of 
roots  of  monomials. 


EVOLUTION.  151 

Rule  . — Extract  the  root  of  the  numeral  coefficients,  and 
divide  the  exponent  of  each  letter  hy  the  index  of  the  root. 

EXAMPLES. 

1 .  What  is  the  second  root  of  SaVj/^  ?     ,  .  Ans.  2>ax^y^. 

2.  What  is  the     third  root  of  SaV  ?  .     .  .  Ans.      9,a^y. 

3.  Whatis  the/owr^A  rootof  81aV2?     .  .  Ans.      2,a^. 

4.  What  is  the     fifth  root  of  32aVV^  ?  •  ^^«-  2a.tY- 

For  illustration,  we  will  observe  that  this  last  example  requires 
us  to  find  five  equal  factors,  which,  when  multiplied  together,  will 
produce  32 ;  and^Zve  equal  factors,  which,  when  multiplied  together, 
will  produce  a^,five  equal  factors,  which,  when  multiplied  together, 
will  produce  x^^ ,  and  ^re  equal  factors,  which,  when  multiplied 
together,  will  produce  y^^. 

Now,  a*  shows  five  equal  factors,  each  equal  to  a  ;  therefore,  a 
is  one  of  the  factors  required.  In  the  same  manner  ar'"  shows 
ten  equal  factors,  and  the  product  of  two  of  these,  or  x^  is  one  of 
the  five  equal  factors  required.  In  the  same  manner  y^  is  another 
of  the  equal  factors  ;  and  there  is  no  trouble  in  finding  any  root 
of  any  hteral  monomial  quantity;  for  all  we  have  to  do  is  to 
divide  its  exponent  (whatever  it  may  be)  by  the  index  of  the 
proposed  root.  But  when  the  factor  is  a  numeral,  like  32,  we  can 
find  the  factor  only  by  trial.  Sometimes  no  such  factor  as  the 
one  required  exists  ;  in  such  cases  we  consider  the  number  as  a 
letter  with  1  understood  for  its  exponent,  and  then  divide  such 
exponent  by  the  index  of  the  root.  For  example,  the  fifth  root 
of  32  is  (32)^;  but  this  is  only  an  indication  of  the  root  or  fac- 
tor, not  an  actual  discovery  of  it.  Take  particular  notice  of  the 
following  examples : 

L    3. 

6.  What  is  the     third  root  oi  1  a^x^  1      .     .     Ans.  I'^a^x. 
The  number  7  is  here  regarded  as  a  letter. 

6.  What  is  the  second  root  of  20aaj?     Ans.  ±(20)20^. 

7.  What  is  the  fourth  root  of  16aV  ? 

Ans.  — 2aa:',  or  2aa;'. 

8.  What  is  the  square  root  of  36aV  ^    •     •    ^^^-  i^ay^. 


li^  ELEME^'TARY    ALGEBRA. 

(Art.  78.)  The  even  roots  of  algebraic  quantities  may- 
be taken  with  the  double  sign,  as  indicating  either  plus  or 
minus,  for  either  quantity  will  give  the  same  square,  and  we 
may  not  know  which  of  them  produced  the  power,  (Art  70). 
For  example,  the  square  root  of  16  may  be  either  -[-4  or — 4, 
for  either  of  them,  when  multiplied  by  itself,  will  produce  16. 

The  cube  root  of  a  plus  quantity  is  always  plus,  and  the 
cube  root  of  a  minus  quantity  is  always  minus.  For  -i-2a 
cubed,  gives  -f-8a^,  and  — 2a  cubed,  gives  -—8a'',  and  a  may 
represent  any  quantity  whatever. 

9.  What  is  the  fourth  root  of  Sla^iV*  ?  Ans.  itSaJV. 
10.  What  is  the  third  root  of  — 27a^V  ?  Ans.  — 3a%. 
U.  What  is  the    third  root  of  I6a^?  Ans.  2a(2a)'^. 

In  this  example  it  is  obvious  that  there  are  no  three  equal 
factors,  which,  when  multiplied  together,  will  produce  16,  and 
there  are  no  three  equal  factors  expressed  in  entire  quantities 

L    1 

that  will  produce  a'*.  Therefore,  we  must  write  (16)^ a^  for 
the  answer.  But  this  is  only  indicating  the  operation,  not 
performing  it,  and  we  have  no  clearer  idea  of  the  result  now 
than  at  first.  However,  to  see  what  can  be  done,  we  will 
separate  16a!^  into  the  two  factors  (8a'')(2a).  The  first  of 
these  is  a  complete  third  power,  and  the  other  is  not ;  but  the 
third  root  of  the  whole  is  the  third  root  of  the  two  factors 

written  together  as  a  product;  that  is,  2a(2a)=*,  and  this  is 
all  we  can  do  to  reduce  or  simplify  it. 

12.  What  is  the  second  root  of  20aV  ?    Ans.  ±.2ax{5x)K 

All  the  square  factors  in  this  are  4a?x^,  the  other  fjictors 
are  5x.  We  can  take  the  second  root  of  the  square  factors, 
and  of  the  others  we  cannot.     In  relation  to  them  we  can  only 

indicate  the  root.     Therefore,  the  whole  root  is  d^2ax(5x)'^. 


13.  What  is  the  second  root  of  75  ?     . 

14.  What  is  the  second  root  of  9Qa^x  ? 

15.  What  is  the    third  root  of  32a^  ?  . 

16.  What  is  the    third  root  of  24a";J^  ? 

17.  What  is  the    third  root  of  27a^  ?  . 

18.  What  is  the  third  root  of  19a  ?    . 


EVOLUTION.  16» 

Ans.  ±5(3)^. 

Ans.  ±7a(2x)^. 

Ans.  2a(4)^. 


Ans.  2a{3x'^Y\ 

.    Ans.  3a. 

,Ans.  (19a)  3. 


(Art.  78.)  By  comparing  examples  17  and  18,  we  perceive 
that  some  monomials  have  such  roots  (or  what  is  the  same 
thing),  such  equal  factors  as  may  be  required,  and  some  have 
not.  When  no  such  factors  exist,  all  we  can  do  is  to  indicate 
an  operation,  to  be  performed  as  example  18.  So  it  is  with 
polynomials — some  may  have  equal  factors,  and  others  not. 
When  equal  factors  do  exist  in  any  polynomial,  they  are  com- 
monly apparent  to  any  one  who  has  had  a  little  experience  in 
raising  roots  to  powers  as  explained  in  Articles  73,  74,  and 
75.  For  instance,  any  one  can  perceive  that  the  polynomial 
a'-\-2ab-\-P  has  two  equal  factors,  each  equal  to  (a-\-b)  ;  and 
after  a  little  more  observation  we  can  perceive  that  the  poly- 
nomial a^-\-3x^y-{-Sxi/-]ry^  has  three  equal  factors,  each  equal 
to  (x-\-y),  or  perceive  that  (x-^ry)  is  its  third  root. 

It  is  only  regidar  polynomials  that  have  equal  factors,  and 
it  is  only  by  observing  how  the  powers  are  formed  by  multipli- 
cation that  we  can  determine 

HOW  TO  EXTRACT  ROOTS  OF  POLYNOMIALS. 

On  the  supposition  that  we  know  that  the  square  root  of 

the  polynomial 

a^-^2ab-\-h\  is  (a-fi), 

we  propose  to  extract  it  out  of  the  polynomial  itself. 

We  know  that  a^,  the  first  term,  must  have  been  formed  by 
the  multiplication  of  a  into  itself,  therefore,  a  must  be  part  of 
the  root  sought. 


154  ELEMENTARY  ALGEBRA. 

The  next  term  is  2aXb,  that  is  tidce  tlie  root  of  the  first  term 
into  the  second  term  of  the  root.  Hence,  if  we  divide  the 
second  term  of  the  square  by  twice  the  root  of  the  first  term, 
we  shall  obtain  b,  the  second  term  of  the  root,  and  as  b  must 
be  multiplied  into  itself  to  form  a  square,  we  add  b  to  2a,  and 
2a-{-b  we  call  a  divisor. 

OPERATION. 

a'-{-2ab-\-F(a-{-b 

«2 


2a-\-b)2ab-^b^ 
2ab+b^ 


"We  take  a  for  the  first  term  of  the  root,  and  subtract  its 
square  (a^)  from  the  whole  square.  We  then  double  a  and 
divide  2ab  by  2a  and  w^e  find  b,  which  we  place  in  both  the 
divisor  and  quotient.  Then  we  multiply  2a-\-b  by  5,  and  we 
have  2ab-\-b^,  to  subtract  from  the  tAVO  remaining  terms  of 
the  square,  and  in  this  case  nothing  remains. 

Again,  let  us  take  a-\-b-\-c,  and  square  it.     We  shall  find 

its  square  to  be 

a'-}-2ab-{-b''-\-2ac-\-2bc-\-c' 

a^+2ab-\-b''-{-2ac-\'2bc-{-c\a-\-b-\-c 


2a-\-b         2ab+U' 
2ab-^P 


2a-r2b-^c  2ac-{-2bc-\-cr' 

2ac-\-2bc-\-i^ 


By  operating  as  before,  we  find  the  first  two  terms  of  the 
root  to  be  a-\-b,  and  a  remainder  of  2ac-\-2bc-\-(^.  Double 
the  root  already  found,  and  we  have  2a-\-2b  for  a  partial 
divisor.  Divide  the  first  term  of  the  remainder  2ac  by  2oy 
and  we  have  c  for  the  third  term  of  the  root,  which  must  be 
added  to  2a-\-2b  to  complete  the  divisor.     Multiply  the  divisor 


EVOLUTION.  155 

by  the  last  term  of  the  root,  and  set  the  product  under  the 
three  terms  last  brought  down,  and  we  have  no  remainder. 

Again,  let  us  take  a+^+c  to  square ;  but  before  we  square 
it,  let  the  single  letter  s=a-{-b. 

Then  we  shall  have  s-\-c  to  square,  which  produces 

To  take  the  square  root  of  this,  we  repeat  the  first  opera- 
tion, and  thus  the  root  of  any  quantity  can  be  brought  into 
a  binomial,  and  the  rule  for  a  binomial  root  will  answer  for  a 
root  containing  any  number  of  terms  by  considering  the  root 
already/  found,  however  great,  as  one  term. 

Hence,  the  following  rule  to  extract  the  square  root  of  a  com- 
pound quantity. 

Rule  . — Arrange  the  terms  according  to  the  powers  of  some 
letter,  beginning  vnth  the  highest,  and  set  the  square  root  of  the 
first  term  in  the  quotient. 

Subtract  the  square  of  the  root  thus  found  from  the  first  tetTn, 
and  bring  doion  the  next  two  tei-ms  for  a  dividend. 

Divide  the  first  term  of  the  dividend  by  double  of  the  root 
already  found,  and  set  the  result  both  in  the  root  and  in  the 
divisor. 

Multiply  the  divisor,  thus  completed,  by  the  term  of  the  root 
last  found,  and  subtract  the  product  from  the  dividend,  and  so  on. 

EXAMPLES. 

1 .  What  is  the  square  root  of 

a^-\-4a'b-\-4b''—4a^-'8b-\-4(a^-\-2b—2 

«4 


2a'4"25       )4a'b-\-4P 
4a^^4b^ 


2a^+4b—2  _4a2_86-f-4 


1&6  ELEMENTARY    ALGEBRA. 

2.  Wliat  is  the  square  root  of   1 — 4b-\-4b^-{-2^ — 4^;y+?/2? 

Ans.  1 — 26-ry. 

3.  What  is  the  square  root  of  4x^—4x^-\'\Sx'^—6x-{-9  ? 

Ans.  2x^ — ^ar-f-S. 

4.  What  is  the  square  root  of  4x^—163^-\-24x^ — \6x-\-4  ? 

Ans.  2aP — 4x-\-2. 

5.  What  is  the  square  root  of     IGx^  -f  24a^  +  892:=  +  60a; 
100?  .  A71S.  4x^-i-3x-hy0. 

6.  What  is  the  square  root  of  4x'^^\6a^-^dx'^-{-16x-{-4  ? 

Ans.  2x^ — 4x — 2 

7.  What  is  the  square  root  of  x^ -{-2iy-]-y^ -]-Gxz-\-Gi/z 
+92:2?  -     jns.  x+7/-^3z. 

8.  What  is  the  square  root  of  a^ — ab-\-lb^  ?     Ans.  a — ^b 

a^  b^ 

9.  What  is  the  square  root  of  -r^ — 2-[— j? 


b^ 


.        a     b         b     a 

Ans. or -. 

0     a        a     b 

2.  J     JL  2. 

10.  What  is  the  square  root  of  x"^ — 2x'^y^-\-y^  ? 

.  L  1  1         J. 

Ans.  x^ — y^  or  y^ — x^. 

(Art.  79.)  Every  square  root  will  be  equally  a  root  if  we 
change  the  sign  of  all  the  terms.  In  the  first  example,  for 
instance,  the  root  may  be  taken  — a' — 2b-\-2,  as  well  as 
a^-\-2b — 2,  for  either  one  of  these  quantities,  by  squaring, 
will  produce  the  given  square.  Also,  observe  that  every 
square  consisting  of  three  terms  only,  has  a  binomial  root. 

Algebraic  squares  may  be  taken  for  formulas,  correspond- 
ing to  numeral  squares,  and  their  roots  may  be  extracted  in 
the  same  way,  and  by  the  same  rule. 

For  example,  a-{-b  squared  is  a^-{-2ab-\-b^,  and  to  apply 
this  to  numerals,  suppose  a =40,  and  5=7. 


EVOLUTION.  157 

Then  the  square  of  40  is      a'=1600 

2(z5=  560 

1?^     49 


Therefore,     ....   (47)2=2209 

Now  the  necessary  divisions  of  this  square  number,  2209, 
are  not  visible,  and  the  chief  difficulty  in  discovering  the  root 
is  to  make  these  separations. 

The  first  observation  to  make  is,  that  the  square  of  10  is 
100,  of  100  is  10000,  and  so  on.  Hence,  the  square  root  of 
any  square  number  less  than  100,  consists  of  one  figure,  and 
of  any  square  number  over  100  and  less  than  10000,  of  two 
figures,  and .  so  on.  Every  two  places  in  a  power  demanding 
one  place  in  its  root. 

Hence,  to  find  the  number  of  places  or  figures  in  a  root, 
we  must  separate  the  power  into  periods  of  two  figures,  begin- 
ning at  the  unit's  place.  For  example,  let  us  require  the 
square  root  of  22  09.  Here  are  two  periods  indicating  two 
places  in  the  root,  corresponding  to  tens  and  units.  The 
greatest  square  in  22  is  16,  its  root  is  4,  or  4  tens  =40. 
Hence,  a =40. 

22  09(40+7=47 
a2=l6  00 

2a+J=804-7=87    )6  09 
6  09 


Then  2a=80,  which  we  use  as  a  divisor  for  609,  and  find 
it  is  contained  7  times.  The  7  is  taken  as  the  ■^lue  of  h,  and 
2a-{-b,  the  complete  divisor,  is  87,  which,  multiplied  by  7,  gives 
the  two  last  terms  of  the  binomial  square.  2ab-{-b^~560-\-49 
=609,  and  the  entire  root,  40-|-7=47,  is  found. 

Arithmetically,  a  may  be  taken  as  4  in  place  of  40,  and 
1600  as  16,  the  place  occupied  by  the  16  makes  it  16 
hundred,  and  the  ciphers  are  superflous.     Also,  2a  may  be 


394 

369 

25  24 

25  24 

158  ELEMENTARY  ALGEBRA. 

considered  8  in  place  of  80,  and  8  in  60  (not  in  609)  is 
contained  7  times,  &c. 

If  the  square  consists  of  more  than  two  periods,  treat  it  as 
two,  and  obtain  the  two  superior  figures  of  the  root,  and  when 
obtained,  bring  down  another  period  to  the  remainder,  and 
consider  the  root  already  obtained  as  one  quantity,  or  one 
figure. 

For  another  example,  let  the  square  root  of  399424  be 
extracted. 

39  94  24(632 
36 

123 


1262 


In  this  example,  if  we  disregard  the  local  value  of  the 
figures,  we  have  a=6,  2a=12,  and  12  in  39,  3  times,  which 
gives  5=3.  Afterward  we  suppose  a=63,  and  2a=:126, 
126  in  252,  2  times,  or  the  second  value  of  5=2.  In  the 
same  manner,  we  would  repeat  the  formula  of  a  binomial 
square  as  many  times  as  we  have  periods. 

EXERCISES  FOR  PRACTICE. 

1.  What  is  the  square  root  of  8836  ?    .     .     .     Ans.     94. 

2.  What  is  the  square  root  of  106929?      .     .     Ans,  327. 

3.  What  is  the  square  root  of  4782969  ?    .     .  A7is.  2187. 

4.  What  is  the  square  root  of  43046721  ?  .     .  Ans.  Q5Q\. 

5.  What  is  the  square  root  of  387420489  ?      Ans.  19683. 

When  there  are  whole  numbers  and  decimals,  point  off 
periods  both  ways  from  the  decimal  point,  and  make  the  deci- 
mal places  even,  by  annexing  ciphers  when  necessary,  extend- 
ing the  decimal  as  far  as  desired.  When  there  are  decimals 
only,  commence  pointing  off  from  the  decimal  point. 


( 


EVOLUTION.  159 

EXAMPLES. 

1.  What  is  the  square  root  of  10.4976  ?    .     .     Ans,  3.24. 

2.  What  is  the  square  root  of  3271.4207?    Ans.  57.19+. 

3.  What  is  the  square  root  of  4795.25731  ? 

Ans.  69.247+. 

4.  What  is  the  square  root  of  .0030  ?        .     .       Ans.  .06. 

5.  What  is  the  square  root  of  .00032754  ? 

Ans.  .01809+. 

6.  What  is  the  square  root  of  .00103041  ?        Ans.  .0321. 

As  the  square  of  any  quantity  is  the  quantity  multiplied 
by  itself,  and  the  product  of  -  by  -  (Art.  64)  is  j^;  hence,  to 

take  the  square  root  of  a  fraction,  we  must  extract  the  square 
/  root  of  both  numerator  and  depominator. 

A  fraction  may  be  equal  to  a  square,  and  the  terms,  as  given, 
not  square  numbers ;  such  may  be  reduced  to  square  numbers. 

EXAMPLES. 

What  is  the  square  root  of  -^-^j  ? 

Observe  tVi"— e  I •                    Hence,  the  square  root  is  f . 

1 .  What  is  the  square  root  of  /2V  ?      •     •     •     •     Ans.  |. 

2.  What  is  the  square  root  of  i||  ?      .     .     .     .     Ans.  f . 

3.  What  is  the  square  root  of  f  s||  ?    .     .     .     .     Ans.  f . 

4.  What  is  the  square  root  of  |||A  ?    .  ,  .     .     .     Ans.  |. 
When  the  given  fractions  cannot  be  reduced  to  square 

terms,  reduce  the  value  to  a  decimal,  and  extract  the  root,  as 
in  the  last  article. 

TO  EXTRACT  THE  CUBE  ROOT  OF  COMPOUND 
QUANTITIES. 

(Art.  80.)  We  may  extract  the  cube  root  in  a  similar 
manner  as  the  square  root,  by  dissecting  or  retracing  the 
combination  of  terms  in  the  formation  of  a  binomial  cube. 

The  cube  of  a-\-b  is  a^-\-'Sa'^b-\-2,alP-\-b'^  (Art.  67).  Now, 
to  extract  the  root,  it  is  evident  we  must  take  the  root  of  the 


160  ELEMENTARY  ALGEBRA. 

first  term  (a'),  and  the  next  term  is  3a'^b.  Three  times  the 
square  of  the  first  letter  or  term  of  the  root  multiplied  by  tJie  9.d 
term  of  the  root. 

Therefore,  to  find  this  second  term  of  the  root,  we  must 
divide  the  second  term  of  the  power  (3a^5)  by  three  times 
the  square  of  the  root  already  found  (a). 

Sa'b 

When  we  can  decide  the  value  of  b,  we  may  obtain  the 
complete  divisor  for  the  remainder,  after  the  cube  of  the  first 
term  is  subtracted,  thus  : 

The  remainder  is      .     3a^b-{-3a^b-\-b^ 

Take  out  the  factor  b,  and  3a?-\-3ab-\-b^  is  the  complete 
divisor  for  the  remainder.  But  this  divisor  contains  b,  the 
very  term  we  wish  to  find  by  means  of  the  divisor ;  hence, 
it  must  be  found  before  the  divisor  can  be  completed.  In 
distinct  algebraic  quantities  there  can  be  no  difficulty,  as 
the  terms  stand  separate,  and  we  find  b  by  dividing  simply 
Sa^b  by  3a^;  but  in  numbers  the  terms  are  mingled  together 
and  b  can  only  be  found  by  trial. 

Again,  the  terms  3a?-\-3ab-\-b^  explain  the  common  arith- 
metical rule,  as  3a^  stands  in  the  place  of  hundreds,  it  corres- 
ponds with  the  words  :  **  Multiply  the  square  of  the  quotient 
by  300,"  *♦  and  the  quotient  by  30,"  (3a),  &c. 

By  inspecting  the  various  powers  of  a-\-b  (Art.  73),  we 
draw  the  following  general  rule  for  the  extraction  of  roots  : 

Rule  . — Arrange  the  terms  a/icording  to  the  po^vers  of  some 
letter;  take  the  required  root  of  the  first  term  and  place  it  in,  the 
quotient ;  subtract  its  corresponding  power  from  the  first  term, 
and  bring  down  the  second  term  for  a  dividend. 

Divide  this  term  by  tuice  the  root  already  found  for  the 
SQUARE  root,  three  times  the  square  of  it  for  the  cube  root,  four 
times  the  third  power  for  the  fourth  root^  ^kc,  and  the  qucHetU 


EVOLUTION.  161 

wUl  be  the  next  term  of  the  root.  Involve  the  whole  of  the  root 
ihvs  found,  to  its  proper  power,  which  subtract  from  the  given 
quantity,  and  divide  the  first  term  of  the  remainder  hy  the  same 
divisor  as  before ;  proceed  in  this  manner  till  the  whole  root  is 
determined. 

EXAMPLES. 

1.  What  is  the  cube  root  of  x^+6a^~403^-\-96x—64  ?       - 
x'-\-6aP—40x^-\-96x—64:  (x^-\-2x—4. 
x^ 


Divisor  Sx'^)  6a:^=  1st  remainder. 

Divisor   Sx*      ) —  12a;'*=2d  remainder. 
a^-J^ex^—40x^+96x—6l 

2.  What  is  the  cube  root  of  27a^4-108a=+144a+64  ? 

Ans.  3a+4. 

3.  What  is  the  cube  root  of  c^—6a^x-h l^ax^—^a^  ? 

Ans.  a — 2x. 

4.  What  is  the  cube  root  of  x^—3a^-\-5sP—2x—l  ? 

Ans.  x^ — X — 1. 
6.  What  is  the  cube  root  of  a^—6a''b-]-nah^--8b'''i 

Ans.  a — 2b. 

3     1 

6.  What  is  the  cube  root  of  •'^+3.r+-+-3  ?     ^^^       .1 


X 


7.  Extract  the  fourth  root  of 

a^+8a^+24a2+32a+16(a-f2 


4a^)       8a^  <fec. 


a^-l-8a''+24a2+32a+16 


(Art.  81.)  To  apply  this  general  rule  to  the  extraction  of 

the  cube  root  of  numbers,  we  must  first  observe  that  the  cube 

of  10  is  1000,  of  100  is  1000000,  <fec.;  ten  times  the  root 

producing  1000  times  the  power,  or  one  cipher  in  the  roa+ 

14 


162  ELEMENTARY   ALGEBRA. 

producing  3  in  the  power;  hence,  any  cube  within  3  places 
of  figures  can  have  only  one  in  its  root,  any  cube  within  6 
places  can  have  only  two  places  in  its  root,  &c.  Therefore, 
we  must  divide  off  the  given  power  into  periods  consisting  of 
three  places,  commencing  at  the  unit.  If  the  power  contains 
decimals,  commence  at  the  unit  place,  and  count  three  places 
each  way,  and  the  number  of  periods  will  indicate  the  number 
of  figures  in  the  root. 

EXAMPLES. 

1.  Required  the  cube  root  of  12812904. 
12  812  904(234 
a=2         a^z=z  8 


Divisor       3^2=  12  )48 

12167  =  (23)^^ 
3(23)2=1587)       6459"(4 

12  812  904=(234)« 

Here,  12  is  contained  in  48,  4  times;  but  it  must  be 
remembered  that  12  is  only  atrial  or  partial  divisor;  when 
completed  it  will  exceed  12,  and  of  course  the  next  figlire  of 
the  root  cannot  exceed  3. 

The  first  figure  in  the  root  was  2.  Then  we  assumed  a=2. 
Afterward  we  found  the  next  figure  must  be  3.  Then  we 
assumed  a=23.  To  have  found  a  succeeding  figure,  had 
there  been  a  remainder,  we  should  have  assumed  a =234, 
&c.,  and  from  it  obtained  a  new  partial  divisor. 

2.  What  is  the  cube  root  of  148877?    . 

3.  What  is  the  cube  root  of  571787  ?    . 

4.  What  is  the  cube  root  of  1367631  ? 
6.  What  is  the  cube  root  of  2048383  ? 

6.  What  is  the  cube  root  of  16581375 1 

7.  What  is  the  cube  root  of  44361864  ? 

8.  What  is  the  cube  root  of  100544625  ? 


.     .     Am. 

53. 

Ans. 

83. 

,     .     Ans. 

111. 

.     .     Ans. 

127. 

.     .     Ans. 

255. 

.     .     Ans. 

354. 

.     .     Am. 

465. 

EVOLUTION.  163 

(Art.  82.)  The  methods  of  dh'cct  extraction  of  the  cube 
root  of  such  numbers  as  have  surd  roots,  are  all  too  tedious 
to  be  much  used,  and  several  eminent  mathematicians  have 
given  more  brief  and  practical  methods  of  approximation. 

One  of  the  most  useful  methods  may  be  investigated  as 
follows : 

Suppose  a  and  a-\-c  two  cube  roots,  c  being  very  small  in 
relation  to  a,  a?  and  a^+Sa^c+Sac^-f-c^  are  the  cubes  of  the 
supposed  roots. 

Now,  if  we  double  the  first  cube  (a^),  and  add  it  to  the 
second,  we  shall  have 

If  we  double  the  second  cube  and  add  it  to  the  first,  we 
shall  have  Sa^+Ga'c-f- 6(2(^4- Sc^ 

As  c  is  a  very  small  fraction  compared  to  a,  the  terms  con- 
taining c^  and  <?  are  very  small  in  relation  to  the  others ;  and 
the  relation  of  these  two  sums  will  not  be  materially  changed 
by  rejecting  those  terms  containing  c^  and  c^,  and  the  sums 
will  then  be  .     .     .     .     Sa^-f  Sa^c 

And       .....     Za^-\-Qa?c 

The  ratio  of  these  terms  is  the  same  as  the  ratio  of  a-j-c  to 

Or  the  ratio  is 14— j—- 

/• 

But  the  ratio  of  the  roots  a  to  a+c,  is  1  +  -. 

a 

Observing  again,  that  c  is  supposed  to  be  very  small  in 

relation  to  a,  the  fractional  parts  of  the  ratios  — r—  and  -  are 
^  a-\-c         a 

both  small,  and  very  near  in  value  to  each  other.  Hence,  we 
have  found  an  operation  on  two  cubes  which  are  near  each 
other  in  magnitude,  that  will  give  a  proportion  very  near  in 
proportion  to  their  roots  ;  and  by  knowing  the  root  of  one  of 
the  cubes,  by  this  ratio  we  can  find  the  other. 


164  ELEMENTARY  ALGEBRA. 

For  example,  let  it  be  required  to  find  the  cube  root  of  28, 
true  to  4  or  5  places  of  decimals.  As  we  wish  to  find  the 
cube  root  of  28,  we  may  assume  that  28  is  a  cube.  27  is  a 
cube  near  in  value  to  28,  and  the  root  of  27  we  know  to  be  3. 

Hence,  a,  in  our  investigation,  corresponds  to  3  in  this 
example,  and  c  is  unknown ;  but  the  cube  of  a+c  is  28,  and 
a^  is  27. 

Then      ...     27        28 
2  2 

64         66 

Add        ...     28         27 

Sums      .     .     .     82    :    83  :  :  3 :  a-\-c  very  nearly. 

Or,  (a-l-c)=Y/ =3.03658-1-,  which  is  the  cube  root  of 
28,  true  to  6  places  of  decimals. 

By  the  laws  of  proportion,  which  we  hope  more  fully  to 
investigate  in  a  subsequent  part  of    his  work,  the  above  pro- 
portion,      .     .     .     .     82  :  83  :  :  a :  a+c, 
may  take  this  change,    82  :    1  :  :a:c 

Hence,  c=j\;  c  being  a  correction  to  the  known  root,  which, 
being  applied,  will  give  the  unknown  or  sought  root. 
From  what  precedes,  we  may  draw  the  following  rule  for 
finding  approximate  cube  roots  : 

Rule  . — Take  the  nearest  rational  cube  to  the  given  number y 
or  assume  a  root  and  cube  it.  Double  this  cube,  and  add  the 
number  to  it ;  also  double  the  number  and  add  the  assumed  cube 
to  it.  Then,  by  proportion,  as  the  first  sum  is  to  the  second, 
so  is  the  known  root  to  the  required  root. 

EXAMPLES. 

1.  What  is  the  approximate  cube  root  of  122  ? 

Ans.  4.95967-f. 
By  the  rule     .     125X2=250         244 
Add      ...  122         125 

372    :    369  :  :  5 :  Ans. 


EVOLUTION.  165 

2.  What  is  tlie  cube  root  of  10  ?      .     .     Ans.  2.1 5U3-\-, 
Assume  2.1  for  the  root,  then  9.261  is  its  cube. 

3.  What  is  the  approximate  cube  root  of  720  ? 

Ans.    8.9628+. 

4.  What  is  the  approximate  cube  root  of  345  ? 

Ans.  7.013574-. 

5.  What  is  the  approximate  cube  root  of  520  ? 

Ans.  8.04145-f. 

6.  What  is  the  approximate  cube  root  of  65? 

Ans.    4.0207+. 

7.  What  is  the  approximate  cube  root  of  16  ? 

The  cube  root  of  8  is  2,  and  of  27  is  3;  therefore  the  cube 
root  of  16  is  between  2  and  3.  Suppose  it  2.5.  The  cube 
of  this  root  is  15.625,  which  shows  that  the  cube  root  of  16 
is  a  little  more  than  2.5,  and  by  the  rule 

31.25         32 
16  15.625 


47.25    :    47.625 :  :  2.5  :  to  the  required  root. 

47.25    :        .375:  :  2.5:. 01984 

Assumed  root  .     .     2.50000 
Correction   .     .     .       .01984 


Approximate  root      2.51984 

We  give  the  last  as  an  example  to  be  followed  in  most 
cases  where  the  root  is  about  midway  between  twointegi-al 
numbers. 

This  rule  may  be  used  with  advantage  to  extract  the  root 
of  perfect  cubes,  when  the  powers  are  very  large. 

EXAMPLE. 

The  number  22.069.810.125  is  a  cube  ;  required  its  root. 
Dividing  this  cube  into  periods,  we  find  that  the  root  must 
contam  4  figures,  and  the  superior  period  is  22,  and  the  cube 


166  ELEMENTARY  ALGEBRA. 

root  of  22  is  near  3,  and  of  course  the  whole  root  near  3000; 
but  it  is  less  than  3000.  Suppose  it  2800,  and  cube  this 
number.  The  cube  is  21952000000,  which,  being  less  than 
the  given  number,  shows  that  our  assumed  root  is  not  large 
enough. 

To  apply  the  rule,  it  will  be  sufficient  to  take  six  superior 
figures  of  the  given  and  assumed  cubes.     Then  by  the  rule, 


219520 
2 

220698 
2 

'\ 

439040 
220698 

441396 
219520 

659738 

:    660916::  2800 
659738 

659738 

:         1178:: 2800 
2800 

942400 
2336 

659738)3298400(5 
32986.90 

Assumed  root, 
Correction,    . 

2800 
6 

True  root,     ,     2805 

The  result  of  the  last  proportion  is  not  exactly  5,  as  will 
be  seen  by  inspecting  the  work ;  the  slight  imperfection  arises 
from  the  rule  being  approximate,  not  perfect. 

When  we  have  cubes,  however,  we  can  always  decide  the 
unit  figure  by  inspection,  and,  in  the  present  example,  the 
unit  figure  in  the  cube  being  5,  the  unit  figure  in  the  root 
must  be  5,  as  no  other  figui-e  when  cubed  will  give  5  in  the 
place  of  units. 

[For  several  other  abbreviations  and  expedients  in  extract' 
ing  cube  root  in  numerals,  see  **  Robinson's  Arithmetic.**] 


EVOLUTION.  167 

To  obtain  the  4tli  root,  we  may  extract  the  square  root  of 
the  square  root.  To  obtain  the  6th  root,  we  may  take  the 
square  root  first,  and  then  the  cube  root  of  that  quantity. 

To  extract  odd  roots  of  high  powers  in  numeral  quantities 
is  very  tedious,  and  of  no  practical  utility ;  we,  therefore,  give 
no  examples. 

(Art.  83.)  It  is  sometimes  necessary  to  multiply  roots 
together  or  to  divide  one  by  another,  and  we  must,  therefore, 
find  rules  for  such  operations. 

For  instance,  I  wish  to  find  the  product  of  the  square  root 
of  3  into  the  square  root  of  12,  and  I  know  not  how  to  find 
it,  unless  I  first  extract  the  square  root  of  3,  and  then  of  12, 
and  multiply  the  two  roots  together.  But  this  would  require 
a  great  amount  of  labor,  and  even  then  it  would  not  be 
done  to  accuracy,  as  no  exact  square  roots  of  either  3  or  12 
exist. 

It  is  possible,  however,  that  the  product  of  these  two  roots 
is  the  same  as  the  square  root  of  the  product  of  3  and  12, 
that  is,  the  square  root  of  36,  which  is  6;  but  how  are  we  to 
demonstrate  whether  this  be  true  or  not  ? 

In  answer  to  this  inquiry,  we  say  let  a  and  h  represent  any 

two  numbers  then  a-  and  h-  will  represent  their  square  roots, 
(Art.  76). 

In  Algebra  we  represent  the  product  of  any  two  quantities 
by  writing  the  quantities  as  factors  with  or  without  tlie 
sign   of  multiplication   between   them.     Thus,    the   product 

L  J.  L      1 

of  X  and  y  is  x*y  or  xy,  so  the  product  of  a^  into  h"^  is  a^'S-; 
but  the  question  is  whether  or  not  that  this  is  the  same  as 

{alf. 

Now  the  product  of  these  two  roots  must  be  some  number. 
Let  that  number  be  indicated  by  P.     Then  we  shall  have 

this  equation       .     .     .      P=:a^h' 


168  ELEMENTARY  ALGEBRA. 

Square  both  members  of  this  equation  (and  we  square  by 
doubling  the  exponent  of  every  factor,  Art.  71),  and  we  have 

Kow,  by  considering  ah  as  a  single  number,  and  extracting 
the  square  root  of  both  members,  we  have 

This  last  equation  answers  the  questioVy  and  we  learn  thai  the 
product  of  the  roots  is  the  same  as  the  root  of  the  product. 

Hence,  (3)^  X  (12)2=(36)2=6. 

What  is  the  product  of  (2)2  by  (8)^ Ans.  4. 

What  is  the  product  of  5(5)2  by  3(8)2.   .   Ans.  15(40)2. 

Hence,  when  we  wish  to  multiply  numbers  together  which 
contain  factors  under  the  same  root,  we  have  the  following 

Rule  . — Multiply  the  rational  parts  together  for  the  rational 
part  of  the  product,  and  the  radical  parts  together  for  the  radical 
part  of  the  product. 

EXAMPLES. 

1.  Multiply  a(/^)2  by  c(c?)2 Ans.  ac{hdy. 

2.  Multiply  3(3)2  by  2(3)^.       ......  Ans.  18. 

3.  Multiply  3(2)2  by  4(8)2 Ans.  48. 

4.  Multiply  2(14)^  by  3(4)3 Ans.  Q{m)^. 

But  in  56  there  is  a  cube  factor  8,  the  other  factor  is  7, 

therefore  the  last  answer  is  6(8)3(7)3=12(7)3. 

5.  Multiply  2(5)^  by  2(10)^ Ans.  20(2)i 

(Art.  84.)  The  roots  to  be  multiplied  together  may  not  be 
the  same — one  may  be  a  square  root,  the  other  a  cube  or  some 
other  root.  In  such  cases,  is  there  any  other  mode  of  express- 
ing the  product  except  by  a  representation  of  the  factors  ? 


EVOLUTION.  169 

For  example,  what  is  the  product  of  a*  by  6*.  Is  there 
another  manner  of  expressing  it  than 

As  in  the  last  article,  the  product  must  be  some  number 
which  we  can  represent  by  P. 

Then F  =:ah^ 

By  squaring,  .  .  .  F^—ah^ 
By  cubing,  ...  .  .  I^z=zc?l^ 
Taking  the  6th  root,     .    P  =i{c^l^)^ 

In  this  manner  we  can  find  the  product  of  other  roots  ;  but 
in  a  work  like  this  it  is  not  important  to  carry  this  subject  to 
any  great  length  ;  we  give,  however,  the  following 

EXAMPLES. 

1.  What  is  the  product  of  a^"  by  a^  ?      .     .  .     Ans.  a*. 

2.  What  is  the  product  of  (6)^  by  (150)^?  .     Ans,  30. 

3.  What  is  the  product  of  (^)^  by  (|)^  ?   .  Ans.  ^(3)*. 

4.  What  is  the  product  of  (2)^  by  (2)^?   .  Ans.  (32)*. 

^Art.  85.)  As  division  is  the  converse  of  multiplication, 
we  may  infer  at  once  from  Art.  83  that  the  quotient  arising 
from  the  division  of  one  root  by  another,  is  the  same  as  the 
root  of  the  quotient;  but  for  greater  clearness  we  had  better 
denote  the  quotient  by  a  letter  and  use  an  equation. 

For  example,  divide  (8)2  by  (2)^,  the  quotient  is  (4)*  or 
2;  but  to  establish  the  principle  of  operation,  let  Q  represent 
the  required  quotient.  Here,  as  in  all  examples  of  division, 
the  product  of  the  divisor  and  quotient  is  equal  to  the  divi- 
dend.    Therefore,  in  this  case  we  must  have 

22  ^=8^ 

By  squaring,    .     .     .     2C*=8 
15 


170  ELEMENTARY  ALGEBRA. 

This  equation  shows  that  to  obtain  the  true  quotient,  we 
must  divide  one  number  hy  the  other  regardless  of  the  root,  and 
then  vrrite  tJie  root  over  the  quotient. 

Thus, §=(4)2 

EXAMPLES. 

1.  Divide  (54)^  by  (6)3 Ans.     92=3. 

2.  Divide  8(72)2  by  2(6)^.  ,     .     .*  •,  .,     ^^5,    4(12)^. 

3.  Divide  3(10)2  by  (15)2 ^^5.        (6)2. 

(15)2^=3(10)2 

15  §2=9X10,  or  §=^=6 

4.  Divide  18  by  2(3)^ Ans.  3(3)^ 

5.  Divide  6a  by  3(a) 2 Ans.  2(a)"^ 

6.  Divide  (160)^  by  (8)^ Ans.  2(5)^ 

7.  Divide  9  by  (27)t Ans.  (3)2 

8.  Divide  1  by  (j^)2 Ans.  (3)2 

9.  Divide  a  by  (a)2 Ans.  (a)2 

(Art.  86.)  When  the  roots  are  different,  t^^e  proceed  on 
the  same  principle,  which  will  be  sufficient  for  every  possible 
case. 

For  example,  divide  72  by  1'^',  the  quotient  must  be  some 
number  which  we  can  represent  by  Q,  and  from  the  equation 

73^=72 

Cubing,  ....  7^5=72 
Squaring,  ....  7^§®=7^ 
Dividing  by  7'      .     .       (2^=7.     Hence,  Q=l^,   Ans. 

2.  Divide  {a^lPd^)^  hy  d^ Ans.  (ah)^. 


SECTION   IV. 


EQUATIONS. 

(Art.  87.)  We  have  thus  far  been  able  to  resolve  only 
simple  equations,  or  equations  of  the  first  degree ;  but  many 
problems  and  many  philosophical  investigations  present  equa- 
tions of  the  second,  third,  and  higher  degrees,  which  may 
demand  a  solution,  as,  for  instance,  the  first  example  of  Art. 
86  incidentally  demanded  the  solution  of  the  equation 

which  is  an  equation  of  the  sixth  degree  ;  but  it  appears  in  so 
simple  a  form  that  there  is  no  mistaking  the  principle  on  which 
its  solution  depends,  and  thus  generally,  When  an  unknown 
quantity  is  involved  to  any  power,  we  find  the  first  power 
(that  is,  the  quantity  itself)  by  extracting  the  corresponding 
root  of  both  members. 

As  in  that  equation  .     .     ^=(7)^ 

In  the  same  manner,  if  a;"=a,  then  x=a^. 

The  converse  of  these  equations  may  often  occur,  that  is, 
the  unknown  quantity  may  appear  under  the  form  of  a  root, 
as  in  the  following  equation : 

Here,  it  is  obvious  that  the  value  of  x  must  be  found  by 
cubing ;  but  if  we  cube  the  first  member  of  the  equation,  we 
must  cube  the  second  to  preserve  equality,  (Ax.  9).     That  is 


172  ELEMENTARY  ALGEBRA. 

(Art.  88.)  From  the  foregoing  observations,  we  draw  the 
following  general  rule  of  operation : 

Rule  . — To  free  a  quantity  from  a  power,  extract  the  corves- 
jponding  root.  To  free  it  from  a  root,  involve  to  the  correspond- 
ing power. 

When  the  unknown  quantity  is  connected  to  a  known  quan- 
tity, and  the  whole  number  a  power  or  root,  the  power  or 
root,  as  the  case  may  be,  is  removed  in  the  same  manner  as 
before. 

Thus (x-]rciy=cip 

The  value  of  x  is  found,  by  first  taking  the  cube  root  of 
both  numbers  and  afterward  transposing  a. 

Again      ....  (2a;-j-c)2'=a 

Here,  after  squaring  both  members,  we  have 

9.x-]rc=a?y  a  simple  equation. 

(Art.  89.)  The  equations  that  appear  in  Articles  87  and 
88,  and  all  other  equations  of  like  kind  where  the  unknown 
quantity  is  raised  to  a  complete  power,  or  is  under  some  one 
particular  root,  are  called 

PURE    EQUATIONS. 

Thus,  the  equation  ax^=by  or,  which  is  the  same  thing, 

«^=-  is  a  pure  equation,  because  the  power  of  the  unknown 

quantity  is  complete;  but  the  equation  x^-^-hx^c  is  not  a  pure 
equation,  because  it  contains  different  powers  of  the  unknown 
quantity. 

The  equation  {^x-\-a)^=.c  is  a  pure  equation ;  but  the  equa- 
tion sc^-\-x^c  is  an  impure  equation,  because  it  contains  no 
complete  power  of  the  unknown  quantity. 

Again,  x^-\-^ax-\-(i?=c-\-h  is  a  pure  equation,  because  the 
first  member  is  a  complete  power  of  (xAro),  and  {x-\-a)  may 
be  represented  by  y,  then  y^:=^c-\'hy  obviously  a  pure  equation. 


EQUATIONS.  173 

There  is  no  difficulty  in  resolving  pure  equations  as  we 
have  already  seen,  for  all  we  have  to  do  is  to  apply  the  rule 
expressed  in  Art.  88;  but  impure  equations  in  the  higher 
degrees,  -present  serious  di^cuUies  ;  and  even  equations  of  the 
second  degree,  when  impure,  compel  us  to  complete  the  power 
before  we  can  solve  the  equations.  Equations  of  the  second 
degree  can  be  represented  by  a  geometrical  square ;  and  when 
the  equation  is  pure,  the  square  corresponding  to  the  first 
member  is  complete,  and  when  impure,  it  must  be  completed, 
and  the  necessary  operation  is  very  appropriately  called 

COMPLETING    THE    SQUARE 

But  before  we  go  into  the  investigation  of  completing  a 
square,  we  will  give  some  examples  to  exercise  the  learner  in 
resolving  pure  equations. 

EXAMPLES.  • 


1.  Given  ^4-\-(x—2)^==3,  to  find  x.    .     »     Ans.  ar==27. 
To  remove  the  first  radical  sign,  we  square  both  members, 
then     ....     4-{-(a;— 2)2=9 

Dropping  4  from  both  members,  and  then  squaring,  we 


find a;— 2=25 


2.  Given  x — ^x'^-\-6= — 2,  to  find  x.     .     .     Ans.     x—^. 
Transpose  x  for  the  purpose  of  having  the  quantity  under 
the  radical  stand  alone  as  one  member  of  the  equation. 


Thus,    ....  —JaP-hG^^—x 

Now,  by  squaring,  the  radical  sign  will  disappear ;  but  if 
any  other  quantity  were  joined  to  this  by  +  or  — ,  the  radical 
could  not  disappear  in  the  square. 

The  square  is      .     .     x'^-{-6=4 — ix-^-aP 


3.  Given  x-\-JsP — 7=7,  to  find  x.      ,     ,     ,     Ans.  ar=4. 


4.  Given  Jx-jr^2=2-{-Jx,  to  ^nd  x.      .     .     Ans.  x=4. 


fit  ELEMENTARY  ALGEBRA. 

t  N.  B.  No  rules  can  be  given  that  will  meet  every  case,  for 
the  combination  of  quantities  is  too  various.  The  pupil  must 
depend  mainly  on  general  principles  and  his  own  practical 
experience. 

5.  Given  2-\-(3x)^=:j5x-\-4,  to  find  x.   .     Am.  a;=12. 

/  202;2 — 9  \2         J. 

6.  Given    f  — j    =x^,  to  find  x.      .     Ans.  x=  |. 

x^ 

7.  Given  3a;2— 29=--j-510,  tofinda;.       .     Ans    a:=14. 

8.  Given  x-\-2='j4-\-xj64-\-x'^,  to  find  x.    Ans.  x—  6. 

9.  Given  x — ^Jx=Jx'^ — x,  to  find  x,  .     .     Ans.  x=\^, 

10.  Given  xJa?-\-a^=a? — x"^,    to  find  x.       Ans.  x=aj^. 

11.  Given  Jx— 32=16— Jx,  to  find  x.      .     Ans.  x=Sl. 

For  the  sake  of  brevity,  put  a=l6,  then  the  last  equation 
will  be  Jx — 2a=a — Jx.  At  the  conclusion  resubstitute  the 
value  of  a. 


12.  Given  Jx — 16=8 — ^a:,  to  find  a;.  .     .     Ans.  x=25) 

13.  Given  a;^ — aa:^=a:,  to  find  a:.    .     .     .      Ans.  x=- . 

1 — a 


14.  Givena;i-|-^/3+a;= — r=:,  to  find  a:.   .     Ans.  x=   1. 

PROBLEMS  PRODUCING  PURE  EQUATIONS. 

(Art.  90.)  In  solving  problems,  it  often  depends  on  the 
manner  or  means  of  notation  we  employ,  whether  the  equation 
comes  out  simple  or  complex,  or  whether  it  is  a  pure  equation 
or  a  common  quadratic.     For  example, 

1.  Find  two  numbers,  whose  difference  is  6,  and  their  produa 
40.  Ans.  4  and  10. 

If  we  represent  the  least  number  by  x. 

Then  the  greater  number  must  be      x -\-Q 

Their  product  is x^-{-&x 

But,  by  the  problem  this  product  is  40. 


EQUATIONS.  175 

Therefore,  a:^-|-6^'=40;  but  as  yet,  we  cannot  solve  this 
equation,  because  it  contains  two  separate  powers  of  x;  we 
know  not  what  to  do  with  it.  But  in  truth,  the  problem  in 
itself -is  so  simple  we  should  be  able  to  solve  it,  and  must  do 
so  by  some  artifice  or  other. 

After  some  reflection,  we  conclude  to 

Put      .     X — 3  to  represent  the  least  number. 

Then    .     a:-|-3  will  represent  the  greatest  number. 

Product  a^ — 9=40,  a  pure  equation  giving  dr7  for  the 
value  of  X.     (See  i^t.  64). 

The  reason  of  taking  the  double  sign  is  found  in  Art.  78. 

If  we  take  -{-1,  then  the  least  number  is  4  and  the  greater 
10.  If  we  take  — 7,  the  numbers  are  — 10  and  — 4;  but  as 
there  are  really.no  such  numbers  as  — 10  and  — 4,  we  take 
only  +7  from  the  answer  to  x. 

The  solving  of  this  problem  shows  us  how  to  solve  any 
equation  in  the  form  of 

a^-\-ax=b 

Consider  the  first. member  as  the  product  of  x,  and  .r-|-a, 
the  difference  of  theSe  two  factors  is  a. 


Put  .     ,     .     .     .     .     a:=^y--  (1) 

■  fc-\-a='y-\-^  (2) 


Product,     .     .      a;'+aa;=y2— --=5         ^3^ 
Hence,       ....  f=b+~  (4) 


And y==±\f*+-         (5) 


■M 


176  ELEMENTARY  ALGEBRA. 

The  value  af  y,  as  determined  in  equation  (5),  put  in  equa- 
tion (1),  gives 


(Art.  91).  The  student  will  perceive  that  Article  90  is  a 
digression,  but  one  that  should  be  pardoned.  We  now  con- 
tinue our  problems  in  pure  equations,  and  if  any  problem 
does  not  produce  such  an  equation,  it  will  be  because  the 
notation  designed  by  the  author  has  not  been  taken. 

2.  The  sum  of  two  numbers  is  6,  and  the  sum  of  their 
cubes  is  72.     What  are  the  numbers  ?  Ans.  2  and  4. 

Let    S-\-x=  the  greater. 
And  3 — x=:  the  less. 

3.  Divide  the  number  56  into  two  such  parts,  that  their 
product  shall  be  640.  Ans.  40  and  16. 

Let  28+a;=  the  greater, 
28 — x:=  the  less. 

4.  A  and  B  distributed  1200  dollars  each,  among  a  certain 
number  of  persons.  A  relieved  40  persons  more  than  B,  and 
B  gave  to  each  individual  5  dollars  more  than  A.  How 
many  were  relieved  by  A  and  B  ?  Ans.  A,  120;  B,  80. 

Let    a:-f-20=  the  number  relieved  by  A. 
And  X — 20=  the  number  relieved  by  B. 

mi.  1200   ,  "        1200 

Then       ....       -—-+5= — -- 

ar-|-20  ar— -20 

T^•A'\^'  240     ,  ^        240 

Dividmg  by  6,  gives  — ^  i  ^ 


a?-[-20  a;— 20 

To  avoid  numeral  multiplication  and  division,  put  <i^=20, 
and  5=240.     Then  the  equation  becomes 

x-\-a  X — a  f 


EQUATIONS.  177 

6.  Find  a  number,  such  that  one-third  of  it  multiplied  by 
one-fourth,  shall  produce  108.  Ans.  36. 

6.  What  number  is  that  whose  square  plus  18  shall  be 
equal  to  half  its  square  plus  301  ?  Jins.  5. 

7.  What  two  numbers  are  those  which  are  to  each  other  as 
5  to  6,  and  the  difference  of  whose  squares  is  44  ? 

Am.  10  and  12. 
'Lei6x=  the  greater,  and  5x=  the  less. 

8.  What  two  numbers  are  those  which  are  to  each  other  as 
3  to  4,  and  the  diflference  of  whose  squares  is  28  ? 

Ans.  6  and  8. 

9.  What  two  numbers  are  those  whose  product  is  144,  and 
the  quotient  of  the  greater  by  the  less  is  1 6  ? 

Ans.  48  and  3. 

10.  The  length  of  a  lot  of  land  is  to  its  breadth  as  9  to  5, 
and  it  contained  405  square  feet.  Required  the  length  and 
breadth  in  feet.  Ans.  27  and  1 5. 

11.  What  two  numbers  are  those  whose  difference  is  to  the 
greater  as  2  to  9,  and  the  difference  of  whose  squares  is  128  ? 

A71S.  18  and  14. 

12.  Find  two  numbers  in  the  proportion  of  ^  to  |,  the  sum 
of  whose  squares  shall  make  225  ?  Ans.  9  and  12. 

The  thoughtful  student  will  not  use  the  fractional  numbers 
■i-  and  |;  but  he  will  use  whole  numbers  in  the  same  propor- 
tion. Let  this  observation  apply  to  other  problems  as  well 
as  to  this  one. 

13.  There  is  a  rectangular  field,  whose  breadth  is  |  of  the 
length.  After  laying  out  ^  of  the  whole  ground  for  a  garden, 
it  was  found  that  there  were  left  625  square  rods  for  mowing. 
Required  the  length  and  breadth  of  the  field. 

Ans.  Length,  30  rods ;  breadth,  25. 

14.  Two  men  talking  of  their  ages,  one  said  that  he  was 
94  years  old.     Then,  replied  the  younger,  the  sum  of  your 


178  ELEMENTARY  ALGEBRA. 

age  and  mine,  multiplied  by  the  difference  between  our  ages, 
will  produce  8512.     What  is  the  age  of  the  younger  ? 

A71S.  18  years. 

15.  A  fisherman  being  asked  how  many  fish  he  had  caught, 
replied,  **If  you  add  14  to  the  number,  the  square  root  of 
the  sum,  diminished  by  8,  will  equal  nothing."  How  many 
had  he  caught  ?  Aiis.  50. 

16.  A  merchant  gains  in  trade  a  sum,  to  which  320  dollars 
bears  the  same  proportion  as  five  times  the  sum  does  to  2500 
dollars.     What  is  the  sum  ?  Ans.  $400. 

17.  What  number  is  that,  the  fourth  part  of  whose  square 
being  subtracted  from  8,  leaves  a  remainder  equal  to  4  ? 

Ans.  4. 

18.  Find  two  numbers,  such  that  the  second  power  of  the 
greater,  multiplied  by  the  less,  produces  448,  and  the  second 
power  of  the  less,  multiplied  by  the  greater,  gives  392. 

Ans  The  numbers  are  8  and  7. 
Let  x=i  the  greater,  ?/=  the  less. 

Then a:2y:=448  (1) 

And 3^2^392  (2) 

The  product  of  (1)  and  (2)  is 

iry=(448)(392)        (3) 
This  equation  indicates  that  there  may  be  cube  factors  in 
448  and  in  392.     Therefore,  we  will  try  to  find  them  by 
dividing  by  8,  the  least  cube  number  above  unity. 
Thus,     ....     8)448  8)392 

8)56  7)49 

Hence,  .     .     ay=(8«8«7)(8-7'7)=83»73     (4) 
Or,    .     .     .       ay=8«7  (5) 

Divide  (1)  by  (5),  and  we  have    x==S.  • 

19.  A  man  purchased  a  field,  whose  length  was  to  its 
breadth  as  8  to  5.     The  number  of  dollars  paid  per  acre  was 


EQUATIONS.  179 

equal  to  the  number  of  rods  in  the  length  of  the  field :  and 
the  number  of  dollars  given  for  the  whole,  was  equal  to  13 
times  the  number  of  rods  round  the  field.  Required  the 
length  and  breadth  of  the  field. 

Ans.  Length,  104;  breadth,  65  rods. 

Let  8a;=  the  length,  and  5x=:  the  breadth, 

40ar'     a^ 
Then     .    .— — =—  =number  of  acres. 
160      4 

And  .     —XSx=2x^=  the  whole  number  of  dollars. 
4 

Again,  Sx-{-5x=:lSx=  half  round  the  field. 

And  13a;*2»13=  thirteen  times  round,  which  is  equal  to 
the  dollars  paid. 

20.  There  is  a  stack  of  hay,  whose  length  is  to  its  breadth 
as  5  to  4,  and  whose  highth  is  to  its  breadth  as  7  to  8.  It  is 
worth  as  many  cents  per  cubic  foot  as  it  is  feet  in  breadth ; 
and  the  whole  is  worth  at  that  rate  224  times  as  many  cents 
as  there  are  square  feet  in  the  bottom.  Required  the  dimen- 
sions of  the  stack. 

Ans.  Length,  20;  breadth,  16;  and  highth  14  feet. 

7x 
Let   5x=  the  length,  4x=  the  breadth,  and  -^=  highth. 

Then       [  5x*4x*  —  J  4x=  cost  in  cents. 
Again,  5x'4x=  square  feet  on  the  bottom; 

Hence,      .     224* 5a?' 4a:=  cost  in  cents; 

Ix 
Therefore,  ^x'4x'—*4x=29.4'bx*4x 

2 

By  striking  out  equal  factors,  we  have 
7a;'2a:=224 

21.  It  is  required  to  divide  the  number  of  14  into  two  such 
parts,  that  the  quotient  of  the  greater  divided  by  the  less, 


180  ELEMENTARY  ALGEBRA. 

may  be  to  the  quotient  of  the  less  divided  by  the  greater,  as 
16:9.  Ans.  The  parts  are  8  and  6. 

Let  x=  the  greater  part.     Then  14 — x=  the  less. 

/J*  1  4 X 

Per  question,   — : :  :  16  :  9. 

1 4 — X        X 

Multiply  extremes  and  means,  and  — = — ^^ 

Clearing  of  fractions,  we  have  9a;^=16(14 — xy. 
By  evolution,      .     3x=4(l4 — ^a;)=4'14 — Ax. 
By  transposition,     7aJ=4*14. 
By  division     .     .       rr=4»2=8,  the  greater  part. 

We  solved  the  last  four  examples  for  the  purpose  of  strongly 
recommending  the  factor  system,  which  has  not  been  prac- 
ticed or  appreciated  half  as  much  as  its  merits  deserve. 

(Art.  92.)  We  will  now  return  to  equations  in  the  form  of 
x^-}-ax=b,  for  if  we  know  how  to  solve  them,  we  need  not  be 
so  particular  in  our  notation  as  we  have  been  in  the  last 
Article. 

As  we  have  before  remarked,  all  these  equations  can  be 
resolved  by  considering  the  first  member  as  the  product  of 
two  factors,  one  of  which  is  x,  the  other  (x-{-a),  and  their 

difference  is  a.     Then  if  we  put  y — -  for  one  factor,  and  y+- 
for  the  other,  we  shall  have  y^ — 7=^'  ^^^  y^=^-{-- 

That  is,  if  to  b,  or  to  its  equal,  x^-jrax,  we  add  — ,  the 
square  of  -  (the  square  of  half  the  coefficient  of  the  first  power 
of  a;),  we  shall  have  y,  that  is,  some  square.  Hence,  x'^-\-a3t 
H —  is  a  square,  and  we  have  the  equation 


EQUATIONS.  181 

Now,  the  first  member  is  a  square  mform;  but  whether  it 

is  a  numerical  square  or  not,  depends  on  C^+j)  being  a 

numerical  square ;  but  whatever  it  is,  we  have,  by  extracting 
the  square  root  of  both  members 


■=-%^4' 


.->< 


We  have  found  this  formula  before,  in  Art.  90. 

(Art.  93.)  An  equation  in  the  form  of  x^ — ax=bf  its  first 
member  may  be  considered  the  product  of  the  two  factors,  x 
and  {x — a),  and  these  two  factors  diflfer  by  a. 

Let y^l=x  (1) 

And y — ~x — a  (2) 

At 

Multiply  (1)  and  (2),  and  we  have 

y^ — --=a;(a; — a)  =5 

Hence, 2^=^+1'  (3) 

Equation  (3)  shows  that  whether  we  have  an  equation  in 
the  form  of  x^-\'ax-=.h,  or  of  o^—ax=.h;  that  is,  whether  ax 
is  either  -plus  or  minus ^  we  make  the  first  member  a  square  by 

the  addition  of  precisely  the  same  quantity  — ,  which  is  the 

square  of  half  the  coefficient  of  x. 


In  other  words,     .     .     .    x^-{-ax-\-— 

4 

And       ...      ...     x^ — ^'^'7 

The  square  root  of  the  first  is 
The  square  root  of  the  second  is 


are  bpth  squares. 


See  Art.  78. 


182 


ELEMENTARY  ALGEBRA. 


Hence,  to  complete  the  square  of  the  first  member  of  any 
equation  in  the  form  of  x^-}-ax=b,  or  of  x^ — ax=b,  or  more 
generally  when  the  exponent  of  the  unknown  quantity  in  the 
first  term  is  double  of  that  in  the  other,  we  have  the  following 

Rule  . — Add  the  sqttare  of  half  the  coefficient  of  the  lowest 
jyower  of  the  unknown  quantity. 


EXAMPLES. 


Complete  the  squares  in  the  followmg  equations : 

N.  B.  We  add  a  quantity  to  the  first  member  to  complete 
its  square,  we  add  the  same  quantity  to  the  second  member 
to  preserve  equality. 


1.  ar»-f  4a;=96.       . 

.     .     Ans. 

a^+4a;-l-4=96+4. 

2.  ar^— 4a;=46.       . 

.     .     Ans. 

a;2_4a;-|_4==49. 

3.  ^2_7^^8.   .     . 

.     .     Ans. 

ar^— 7a:-{-V=8+V. 

4.  a;'*— 2a;2=24.     . 

.     .     Ans. 

a:4_2:c24-l  =24-1-1. 

5.  a;^"— 4ic"=a.      . 

.     .     Ans. 

a:2«_4^n_|_4_a-|-4. 

6.  a^4-6a;=16.       . 

.     .    \Ans. 

ar^4-6a;+9=25. 

7.  ar*— 15a;=— 54. 

.     .     Ans. 

a?—\bx^-^^=^-\^—b^. 

8.  ar'— |ar=ip.    . 

,     .     Ans. 

^=-|^+i=4^+i. 

9.  x'-^x=\.  .     . 

.     .     Ans. 

^^-f+rV.^i+t'^. 

10.  .     ^.=-.    .     . 

.     .     Ans. 

"^     'f'^Ab'^-'d'^'ib-' 

Find  the  values  of  x  in  each  of  the  ten  preceding  equations. 
First  by  extracting  the  square  root  of  both  members. 

.     .     .  Therefore,    a:=8  or  —12. 


1.  a;4-2=dbl0. 

2.  a;— 2=ir7. 


a;=5  or  — 9. 


EQUATIONS. 


183 


3.  .1;— 1=±|.     .     . 

4.  x^—l=dz5.  .     . 

5.  a;™— 2=±(a+4) 

Therefore,     .     . 

6.  x-{-S=diz5.    . 


7.  a;— V  =±f . 

8.  a:— i=±V. 


9.  o:-/^ 


Therefore,  ir=8  or 


a;=(6)2  orV— 4. 


a  {  c  ,   a?  ^ 


ar=">/2-l-7a4-4  or  (2— 7a-|-4)'* 
.     .      Therefore,  a;=2  or  — 8. 
.     .  "  ar=9  or       6. 

.     .  "  a;=7  or  —  V. 

.     .  "  x=\  or 


1 

6* 
I 
2    .  T 


26      I  dT^b''  I 


The  preceding  ten  equations  are  all  prepared  for  completing 
the  square ;  that  is,  the  highest  power  of  the  unknown  quan- 
tity stands  first,  and  is  positive. 

It  is  necessary  that  it  should  be  positive,  because  we  must 
take  its  square  root,  and  there  is  no  square  root  to  a  negative 
quantity.  Therefore,  in  reducing  an  equation  preparatory  to 
completing  its  square,  if  the  highest  power  comes  out  mi7iits, 
make  it  plus,  by  changing  all  the  signs  to  both  members  of 
the  equation. 

*  '•  Example.  Find  the  values  of  x  from  the  following  equation 

36 


2 


x-\-t 


'  Mtfltiplying  by  2,  and  afterward  dropping  2  from  both 

'•     •   '■■  •  72 

•memJbei^  we  have    .     . 


-x=Q — 


a;  4-2 


•/  Clearing  of  fractions, 

'  •  —f—2x==  Bx+ 1 6—72 

Tran^osmg;  8x,  uniting  16  and  — 72,  and  afterward  chang- 
VRg  all- {he  signs,  we  have 

ar'4-10a;=66.     Hence,  x=4,  or  —14 


184  ELEMENTARY  ALGEBRA. 

2.  Find  the  values  of  x  from  the  equation  3x^-\-2x — 9=76. 

Ans.  x=b     or  — 5|. 

X       2,X         X 

3.  Find  a;  from  2:^4--=-- r+H-  Ans.  x=\     or  —2^. 

Zoo 

x^ 

4.  Find  x  from  - — 30-\-x=2x — 22. 

Ans.  ar=8     or  — 4. 

Cu        X 

5.  Find  a;  from  -• — --j-7i=8i.    .     Ans.  a:=l|^  or  — |. 

^      o 

/^  2a; 

6.  Find  a;  from  — — 16=— — 14f.    Ans.  x=3    or  — ^. 

/p        a;-4-3 

7.  Find  x  from  ——=-—;—.       .     ^W5.  a;=12  or  —2. 

2 

8.  Find  a;  from  a; — 1+ :=0.   •     Ans.  x=3    or       2. 

a: — 4 

9.  Find  x  from     J~^= ;f.     .     Ans.  x=36  or     12. 

20        X — 6 

g/j. J      x-\~  1 

10.  Find  X  from  --=— -r— .     .     Ans.  x=4    or  — 4. 

a; — 1      2a;+7 

(Art.  94.)  For  a  more  definite  understanding  of  quadra- 
tics, we  will  solve  and  strictly  examine  the  following  equation : 
a;2-f4a;=60 
Completing  the  square,  then 

ar^+4a;+4=64 
Extracting  the  square  root  of  both  members,  *,• 

And a:-}-2=±8  »■ 

The  reason  of  taking  the  double  sign  to  8  has  been  ^eiVeral 
times  explained.  '^ 

If  we  take  +8,  then      .    «=6  V 

If     .     .     —8,  then      .     a:=— 10 
That  is,  either  +6,  or  — 10  will  verify  the  equation. 

For 6'+4»6=60  *;;  -' 

Also,     .     .   (—10)2— 4'10=60  •  *    *• 

If  ar=6,     .     .     .     .  ar—  6=0         (1)  *      " 

If  ic=— 10      .     .     .  ar4-10=0         (2) 


EQUATIONS.  185 

If  we  multiply  equations  (1)  and  (2)  together,  we  sliall 
have  as  follows : 

X — 6 

a^—6x 
10a;— 60 


a^-^4x—60=0 


As  the  two  factors  are  in  value  equal  to  0,  the  product  of 
the  two  must,  of  course,  equal  0,  and  we  have  the  equation 
as  above.  Transpose  — 60,  and  we  have  x-\-4x=60,  the 
onginal  equation. 

Thus  we  perceive,  that  a  quadratic  equation  may  be  con- 
sidered as  the  product  of  two  simple  equations,  and  the 
values  of  x  in  the  simple  equations  are  said  to  be  roots  of  the 
quadratic,  and  this  view  of  the  subject  gives  the  rationale  of 
the  unknown  quantity  having  two  values. 

This  example  shows  us  how  to  form  an  equation  when 
we  have  the  two  roots  ;  that  is,  gives  us  the  following 

Rule  . — Connect  each  root  with  a  contrary  sign  to  an  un- 
known quantity.  Take  the  product  of  the  two  binomial  factors 
thus  fanned  for  the  first  member  of  the  equation  sought,  and  0 
for  the  other  member. 

EXAMPLES. 

1.  Find  the  equation  which  has  3  and  — 2  for  its  roots. 

Ans.  x^ — x — 6=0. 

2.  Find  the  equation  which  has  5  and  — 9  for  its  roots. 

Ans.  aP-{-4x — 45=0. 

3.  Find  the  equation  which  has  7  and  — 7  for  its  roots. 

Ans.  a;2— 49=0. 

4.  Find  the  equation  which  has  8  and  — 12  for  its  roots. 

Ans.  x^-\-4x—9e=0. 
16 


166  ELEMENTARY    ALGEBRA. 

Let  the  pupil  observe  that  this  last  equation  is  equation  1 
in  Art.  93,  and  he  can  take  the  roots  of  those  ten  equations 
and  deduce  the  equations  again  if  desirable. 

5.  Find  the  equation  which  has  a  and  h  for  its  roots. 

Ans.  x^ — [a-\-b)x-{-ah=^0. 

If  one  of  the  roots  is  negative,  suppose  — a,  the  equation  is 
then ar^+(a — h)x — ah=0 

If  h  is  negative  and  a  positive,  the  equation  is 

^^-{-{b — a)x — ab=0 
If  both  roots  are  negative,  then  the  equation  is 

x''^{a-^h)x^ab=0 

]Sro^,  let  the  pupil  observe  that  the  exponent  of  the  highest 
power  of  the  unknown  quantity  is  2;  and  there  are  two  roots. 
The  coefficient  of  the  first  power  of  the  unknown  quantity  is  the 
algebraic  sum  of  the  two  roots,  with  their  signs  changed;  and 
the  absolute  term,  independent  of  the  unknown  quantity,  is  the 
product  of  the  roots  (the  sign  conforming  to  the  rules  of 
multiplication). 

From  these  observations  we  can  instantly  form  the  equation 
when  the  two  roots  are  given,  without  the  formality  of  going 
through  the  multiplication  ;  for  example. 

Find  the  equation  which  has  7  and  — 9  for  its  roots. 
7— 9 =—2  ;  changed +2,    7(— 9)=— 63 

Hence,  x'^-i-2x — 63=0  is  the  equation. 

(Art.  95.)  When  a  quadratic  equation  is  formed,  or  found, 
or  given,  we  may  consider  it  as  the  product  of  two  binomial 
factors,  and  those  factors  may  be  obvious  to  one  who  fully 
understands  the  subject,  or  any  one  can  find  them  who  can 
resolve  the  equation. 

In  giving  examples  in  factoring  (Art.  26),  we  omitted 
trinomial  quantities  of  the  second  degree.     The  reason  of 


EQUATIONS.      ,  187 

that  omission  must  now  be  perfectly  comprehended  by  the 
careful  student.  We  now  return  to  that  subject,  and  require 
the  factors  composing  the  expression 

Put  the  expression  equal  to  zero,  and  resolve  the  quadratic, 
and  we  shall  find  the  roots  to  be  — 2  and  — 3.  Therefore, 
the  sought  factors  are  (^+2)  and  (x-\-3). 

Find  the  factors  composing  each  of  the  following  express- 
ions. Each  expression  must  be  taken  as  a  quadratic  equation 
presented  for  solution : 

1.  aP—x—20=zO Ans.  (x—5)(x+4). 

2.  a^— 7a4-12.  ........     Ans.  (a— 3)(a— 4). 

3.  a^— 7a— 8 Ans.  (a—S)(a-{-l). 

4.  x'—x—SO Ans.  {x—6){x-\-5). 

5.  x^-\-7x—\8 Ans.  (x-h9)(x—2). 

6.  x^-\-2ax-^a^ Ans.   (x-\-a)(x-\-a). 

7.  x^ — 2ax-\-a^ Ans.   (x — a)(x — a). 

8.  a^ — a^ Ans.  (x-\-a)(x — a). 

9.  x^ — 2a;+4 Ans.   (x — r)(x — r). 

In  this  last  expression  r=\±J — 3,  and  it  not  being  a 
numerical  quantity,  the  roots  are  said  to  be  imaginary/. 

(Art.  96).  When  a  quadratic  equation  is  reduced  to  the 
form  of  x^-\-ax=b,  to  complete  the  square  of  the  first  mem- 
ber we  take  the  half  of  the  coefficient  of  x  to  square,  there- 
fore, it  will  be  more  convenient  to  represent  that  coefficient  by 
2a  in  place  of  a,  and  as  it  may  have  the  negative  as  well  as 
the  positive  sign,  and  as  b  can  be  negative  as  well  as  positive ; 
therefore,  for  a  representation  of  every  variety  of  quadratic 
equations,  we  have  the  four  general  forms. 

x^-\-2ax=b         (1) 

a^^2ax=b         (2) 

x' — 2ax=—b     (3) 

3^'^2ax=:—b    (4) 


188  ELEMENTARY  ALGEBRA. 

A  solution  of  each  of  these  equations  gives  for  the  values 
of  a;  as  follows : 

x=—a±:Jb  +a2  (1) 
x=^-\-a±Jb  +a'  (2) 
x=-^a^J^—b  (3) 

x=^—a±ijc^^  (4) 

The  quantity  x  has  no  conceivable  value  in  equations  (3) 
and  (4)  when  applied  to  any  problem  in  which  h  has  a  greater 
numerical  value  than  a?,  for  the  solution  requires  the  square 
root  of  Ja^ — b,  a  negative  quantity ;  and  there  being  no  square 
roots  to  such  quantities,  we  have  no  conception  of  any  value 
to  X,  and,  of  course,  we  call  the  value  imaginary. 

After  we  reduce  an  equation  to  one  of  the  preceding  forms, 
the  solution  is  only  substituting  particular  values  for  a  and  b; 
but  in  many  cases  it  is  more  easy  to  resolve  the  equation  as 
an  original  one,  than  to  refer  and  substitute  from  the  formula. 

(Art.  97.)  We  may  meet  with  many  quadratic  equations 
that  would  be  very  inconvenient  to  reduce  to  the  form  of 
a?-{-2a^=b;  for  when  reduced  to  that  form,  2a  and  b  may 
both  be  troublesome  fractions. 

Such  equations  may  better  be  left  in  the  form  of 
aoi^-\-bx=^c 
An  equation  in  which  the  known  quantities,  a,  b,  and  c,  are 
all  whole  numbers,  and  prime  to  each  other. 

"VVe  now  desire  to  find  some  method  of  making  the  first 
member  of  this  equation  a  square,  without  making  fractions. 
We  therefore  cannot  divide  by  a,  because  b  is  not  divisible  by 
a,  the  two  letters  being  prime  to  each  other  by  hypothesis. 
But  the  first  term  of  a  binomial  square  is  always  a  square ; 
therefore,  if  we  desire  the  first  member  of  our  equation  to  be 
converted  into  a  binomial  square,  we  must  render  the  first 
term  a  square,  and  we  can  accomplish  this  by  multiplying 
every  term  by  a. 


EQUATIONS.  iP 

The  equation  then  becomes 

d^oi^-\-hax=ca 

Put y=ax 

Then y^A^hy^ca 

Complete  the  square,  by  the  preceding  rule,  and  we  have 

We  are  sure  the  first  member  is  a  square ;  but  one  of  the 
terms  is  fractional,  a  condition  we  wished  to  avoid ;  but  the 
denominator  of  the  fraction  is  4,  a  squarcy  and  a  square  mul- 
tiplied by  a  square  produces  a  square. 

Therefore,  multiply  by  4,  and  we  have  the  equation 

An  equation  in  which  the  first  member  is  a  binomial  square, 
and  not  fractional. 

If  we  return  the  values  of  y  and  y^  this  last  equation 
becomes  4a'a:^-{-4a5a:-f-5^=4ac+5^ 

Compare  this  with  the  primitive  equation 

"We  multiplied  this  equation  first  by  a,  then  by  4,  and  in 
addition  to  this,  we  find  H^  on  both  sides  of  the  rectified  equa- 
tion, h  being  the  coefficient  of  the  first  power  of  the  unknown 
quantity.  From  this  it  is  obvious,  that  to  convert  the  express- 
ion ax^-^-hx  into  a  binomial  square,  we  may  use  the  following 

Rule  . — Multiply  hy  four  times  the  coefficient  of  x^,  and  add 
the  square  of  the  coefficient  of  x. 

To  preserve  equality,  both  sides  of  an  equation  must  be 
multiplied  by  the  same  quantity,  and  the  same  addition  must 
be  made  to  both  sides.  We  operate  on  the  first  member  of 
an  equation  to  make  it  a  square;  we  operate  on  the  second 
member  to  preserve  equality. 


yfb  ELEMENTARY  ALGEBRA. 

EXAMPLES. 

1.  Given  6a;^+4a;=^4,  to  find  the  values  of  x. 

By  the  rule,  we  multiply  4  times  6,  and  add  to  both  mem- 
bers 4K     That  is, 

4«5V+80a;+16=4080+16 

By  extracting  square  root,  we  have 

2«5a:+4=±64,  a:=6  or  — 6f 

By  extracting  the  square  root  of  the  first  member,  the 
second  term  always  disappears;  it  is,  therefore,  not  necessary 
to  compute  it,  and  for  that  reason  we  may  simply  represent  it 
by  a  letter,  as  in  the  following  example  : 

2.  Given  7x^ — 20a?=32,  to  find  the  values  of  x. 
Multiply  by  4  times  7  and  add  20^. 

Then  .     .     4«7V-— ^+400=896+400 

Square  root,      .     2- 7a; — 20= ±36;  hence,  a;=2  or  — |. 

3.  Given  2x^ — 5a:=117,  to  find  the  values  of  x. 

Ans.  x=9  or  — 6^. 

4.  Given  3aP — 5a; =28,    to  find  the  values  of  x. 

Ans.  x=4  or  — J. 
6.  Given  3a;^ — a;=70,      to  find  the  values  of  x. 

Ans.  x=5  or  — y. 

6.  Given  6a;^+4a;=273,  to  find  the  values  of  x. 

Ans.  x=7  or  — 7|. 

7.  Given  2a;^+3a;=65,    to  find  the  values  of  a;. 

An^.  x=5  or  — 6|. 

8.  Given  33:^+53;= 42,    to  find  the  values  of  ar. 

Ans.  x=3  or  — 4'. 

9.  Given  Sa;'— .7a;+ 1 6  =f  1 8 1 .  to  find  a;. 

Ans.  a;=5  or  — 4J. 


EQUATIONS.  i^l 


10.  Given  10a^—8;r+8=320,  to  find  ir. 

Ans.  x=G  or  — 5}. 

11.  Given  3x^+2x=4,  to  find  ar.    .     Ans.  a;=— i-ii^ls. 

12.  Given  5a;2+7a;=7,  to  find  a;.     Ans.  x— — ^^-^.f^J^. 

240  216 

13.  Given |-to= :-,  to  find  a:.      .  Ans.x=.1b. 

X  X — 15 

QUESTIONS 

GIVING    RISE    TO    QUADRATIC    EQUATIONS. 

1.  If  four  times  the  square  of  a  certain  number  be  dimin- 
ished by  twice  the  number  it  "will  leave  a  remainder  of  30. 
What  is  the  number  ?  Ans.  3. 

N.  B.  The  number  3  is  the  only  number  that  will  answer 
the  required  conditions — the  algebraic  expression  — f  will 
also  answer  the  conditions  ;  but  the  expression  is  not  a  num- 
ber in  any  arithmetical  sense. 

2.  A  person  purchased  a  number  of  horses  for  240  dollars. 
If  he  had  obtained  3  more  for  the  same  money,  each  horse 
would  have  cost  him  4  dollars  less.  Required  the  number  of 
horses.  Ans.  12. 

3.  A  grazier  bought  as  many  sheep  as  cost  him  240  dol- 
lars, after  reserving  15  out  of  the  number,  he  sold  the  remain- 
der for  216  dollars,  and  gained  40  cents  a  head  on  the 
number  sold.     How  many  sheep  did  he  purchase  1     Ans.  15. 

(See  equation  13  just  passed  over). 

4.  A  company  dining  at  a  house  of  entertainment,  had  to 
pay  3  dollars  and  50  cents ;  but  before  the  bill  was  presented 
two  of  them  went  away ;  in  consequence  of  which,  those  who 
remained,  had  to  pay  each  20  cents  more  than  if  all  had  been 
present.     How  many  persons  dined  ?  Ans.  7. 

5.  There  is  a  certain  number,  which  being  subtracted  from 
22,  and  the  remainder  multiplied  by  the  number,  the  product 
will  be  117.     What  is  the  number  ?  Ans.  13  or  9. 


192  ELEMENTARY   ALGEBRA. 

6.  In  a  certain  number  of  hours  a  man  traveled  36  miles, 
but  if  he  had  traveled  one  mile  more  per  hour,  he  would  have 
taken  3  hours  less  than  he  did  to  perform  his  journey.  How 
many  miles  did  he  travel  per  hour  ?  Ans.  3  miles. 

7.  A  man  being  asked  how  much  money  he  had  in  his 
purse,  answered,  that  the  square  root  of  the  number  taken 
from  half  the  number  would  give  a  remainder  of  180  dollars. 
How  much  money  had  he  ?  Ans.  $400. 

8.  Divide  100  into  two  such  parts,  that  the  sum  of  their 
square  roots  may  be  14.  Ans.  64  and  36. 

9.  Divide  the  number  14  into  two  such  parts,  that  the  sum 
of  the  squares  of  those  parts  shall  be  100.        Ans.  8  and  6. 

10.  Divide  the  number  a  into  two  such  parts,  that  the  sum 
of  the  squares  of  those  parts  shall  be  b. 

Ans.  ^{a±ij^b—a?). 

11.  It  is  required  to  divide  the  number  24  into  two  such 
parts,  that  their  product  may  be  equal  to  35  times  their 
difference.  Ans.  10  and  14. 

12.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their 
cubes  152.     What  are  the  numbers  ?  Ans.  3  and  5. 

Let     4 — x=  the  less  number. 
And    4+a;=  the  greater  number. 
Then  put  a=4,  cube,  &c. 

(Art.  98.)  In  the  preceding  examples  wc  have  only  con- 
sidered the  resulting  equation  after  all  the  other  unknown 
quantities  have  been  eliminated. 

In  solving  a  problem,  however,  the  operator  may  use  one, 
two,  three,  or  more  unknown  quantities,  and  operate  as  in  sim- 
ple equations,  and  in  eliminating  one  quantity  after  another, 
there  will  result  a  final  equation,  which  may  be  of  the  first, 
second,  Hiird,  or  higher  degree,  according  to  the  conditions  of 
the  problem  and  the  tact  of  the  operator  in  taking  hold  of  the 
matter. 


EQUATIONS.  193 

(Art.  99.)  When  two  independent  equations  are  drawn 
from  a  problem,  if  one  of  them  is  quadratic,  the  other  must 
be  simple,  or  the  resulting  equation  cannot  be  brought  down 
to  the  second  degree,  except  in  rare  cases,  where  the  two 
equations  are  homogeneous  or  are  symmetrical. 

(Art.  100.)  Two  equations  essentially  quadratic,  involving 
two  unknown  quantities,  depend  for  their  solution  on  a  result- 
ing equation  of  the  fourth  degree. 

(Art.  101.)  No  words  will  cover  every  case  of  similar  or 
symmetrical  equations ;  but  as  a  general  thing,  in  similar 
equations  we  may  change  x  to  y,  and  y  to  a;  without  changing 
the  form  of  the  equations  or  falsifying  them. 

Thus, a:+y=^a 

And icy=b 

Are  similar  equations,  and  x-^y=a 

Are  both  similar  and  symmetrical. 

When  equations  are  similar,  we  can  reduce  them  without 
completing  the  square,  as  the  learner  will  discover  by  the  fol- 
lowing examples : 

1.  Given    \  ,    !•  to  find  ar  and  y. 

(       xy=h  S 

Squaring  the  first,     .     .     x^-\-9.xyAry'^=-a^  (1) 

Four  times  the  second,    .     -     Axy       =45  (2) 

Subtracting  (2)  from  (1)  ^— i^H^y^Z^^HJj  ^gj 

Square  root  of  (3),    ,     .     .     .     x — y—dczja? — 46    (4) 
But a;+y=a  (5) 


Add  (4)  to  (5),    .     .     .    ^.     .         2a;=a±Va2^45  (6) 
Take  (4)  from  (5)    ....         Zy=azpja''—4b  (7) 

Dividing  (6)  and  (7)  by  2  and  we  have  the  values  of  x 
and  y,  and  by  reason  of  the  double  sign  each  letter  has  two 
valueff.    When  the  quantity  ^0^—45  is  a  ccjmplete  secoad 
17 


ELEMENTARY   ALGEBRA. 

power,  the  values  of  x  and  y  will  be  rational,  otherwise  they 
will  contain  surds. 

2.  Given  \  ^, T ^9~f     }J\  Mo  find  x  and  y. 

(  a?-\-f=h     (2)  ) 

Squaring  (1),      .     .      x^-\-9.3cy-]ry'^-=o^  (3) 

Subtracting  (2),       ....    2xy==a?—h  (4) 

Taking  (4)  from  (2),    a?—'2xy-\-y'^=9,h—a?  (5) 

Square  root  of  (5),      .     .       x — y=ziztj2b — a^  (6) 
Adding  (6)  and  (1),    .     .     .      2x=a^j2b-~a^ 
Taking  (6)  from  (1),  .     .     .      2y=a:=f.j20—a^ 

3.  Given  ■!     ,      ,     ^      /«{   f  to  find  x  and  y. 

I  a^—f=b     (2)  ) 

Divide  (2)  by  (1),  and  x—y=-.         (3) 
Equation  (1)  and  (3)  will  give  the  values  sought. 

4.  Given  i     ^  ,    o     ,     \^{   !•  to  find  a;  and  y. 

(  ^-{-y^=b    (2)  j 

.Cube(l),     .     ar'-f3a:V+3a^2^_y3^^3         ^3^ 
From  (3)  take  (2),  and  Sx^y-^Sxy=a^-^   (4) 

Or, Sxy(x-\-y)=a^-b 

Divide  (4)  by  (1),  and  .     .     3xy=a^—-  (5) 

Equations  (1)  and  (5)  combined,  will  make  an  example  the 
same  in  forni  as  example  1,  and  may  be  solved  in  the  same 
manner. 

6.  Given  •<     ,       ~_     \J.    y  to  find  a?  and  y. 

The  equation  resulting  from  these  cannot  be  reduced  to  the 
second  degree,  and  we  mention  the  fact  to  save  the  time  of 
the  operator  from  making  useless  trials. 


EQUATIONS 

Divide  (2)  by  (1),  then  we  have 

1 

> 

x'-\-Ty\-f= 

0 

~  a 

(3) 

Squaring  (1),   .     .     a?—2xy-\-y^= 

=a' 

(4) 

Taking  (4)  from  (3),     .     .     3ary= 

b 

= 

a 

a^ 

(5) 

Dividing  (5)  by  (3),      .     .       xy= 

h 
~3a" 

a" 
3 

Adding  (6)  and  (3),  x'-^^ciyVy^^ 

_45 
"3a" 

a^ 
3 

I9S 


(6) 
(7) 

Square  root  of  (7),      .     .     a;+y=d=^r —    (8) 

Combine  (1 )  and  (8)  for  the  values  of  x  and  y. 
To  exercise  in  these  general  principles j  we  give  the  following 
numeral  examples : 

7.  Given  -<  „   ?•  to  find  x  and  y.      Ans.  i       „        ^ 

(     xy=35)  (y=7  or  6. 

8.  Given  ■<  f-  to  find  x  and  y.  Ans.  \ 

i     xy=  42)  ^  iy=7 

«    ^.  (a;+y=1125)    ,     ^    ,  ,  .         (^ 

9.  Given  -<  „       „     , ,    ^  >-  to  fii^d  a;  and  y.     Ans.  ■{ 

-«    ^.  (^ — y=     4)        ^    ,         ,        .         far=5  or — 1, 

10.  Given  ■{  „      ,     ^^^v  tofindicandy. -<4ws.  •< 

(a:3 — yS_-^24)  ^  (y=l  or — 5. 

cjPJrf=l9(x-{-y)l 
i  x—y=3  j 


or  — 6. 
563, 
562. 


11.  Givenf-^^-^^(^+2/)^,^fi^^^^^y. 


12.  Given' 


a;     y      6 
1      1  __31 


x=5  or  — 2, 
y=2  or  — 5. 


^  to  find  X  and  y.  Ans.  \ 


xz=2  or  3, 
3  or  2. 


To     solve  this,  put,      -=  F,  and  -=Q,  then  we  have  an 
example  in  the  form  of  example  2. 


196  ELEMENTARY  ALGEBRA. 


13.  Given  i      ,      '  >•  to  find  a;  and  y.    Ans.  •!       „  ' 

(Art.  102.) .  Equations  are  homogeneous,  when  the  sum  of 

the  exponents  of  the  unknown  quantities  is  the  same  in  every 

term. 

Thus,  2x^—  X2j=e  )  ,  ..        •. 

,  ,  ^  y  are  homogeneous  equations,  because 

the  sum  of  the  exponents  of  x  and  y  is  the  same  in  every 
term ;  that  is,  2. 

Such  equations  may  always  be  resolved  by  putting  one 
unknown  quantity  equal  to  the  other  multiplied  by  a  new 
unknown  factor.  ' 

To  solve  these  equations,  put  x=vi/. 

Substituting  this  value  of  x  in  the  two  equations,  and  they 
become      ....     2i;y — vy^==6  (1) 

And 2f-\-3vf==S  (2) 

^™-(') 2^=2^        (') 

From  (2)    .....     .  f=^^         (4) 

Equating  (3)  and  (4),  clearing  of  fractions  and  reducing, 
we  have     ....      8v'^ — 13?;= 6  (5) 

Thics,  from  every  pair  of  homogeneous  equations,  we  may 
have  a  resulting  equation  of  the  second  degree  in  reference  to  the 
new  factor  introduced. 

Solving  equation  (6),  we  find  z;=2  or  — f . 

Taking  2  for  the  value,  a:=2y,   and  from  equation  (4) 

o 

y^=  — 7--=l.     Hence,  y=d=l,  which  gives  ar=2,  or  — 2. 

The  equations  .    ic^+y^ — x — ^y=78 

xy-\-x-\-y=2>^y  are  both  quadratic, 
and,  therefore,  by  Art.  100,  they  will  produce  a  resulting 
equation  of  the  fourth  degree  ;  but  they  are  also  similar  and 
symmetrical,  and  for  this  reason,  it  is  possible  to  bring  out  a 


EQUATIONS.  197 

resulting  quadratic,  but  no  general  rule  of  operation  can  be 
laid  down,  and  the  operator  must  depend  mainly  on  his  own 
acquired  tact  and  skill. 

To  resolve  these  equations,  we  double  the  second,  and  add 
it  to  the  first,  we  then  have 

The  first  member  of  this  equation  is  obviously  the  same  as 
(a;-{-y)2+(rc+y)  =  156 

For  the  purpose  of  simplification,  put  (a;-{-y)=s. 

Then,  s^+5=i.  1 56,  a  quadratic  equation  in  relation  to  s,  and 
a  solution  gives  s,  ora;+y=12.  This  equation  taken  from 
the  second  of  the  primitive  equations  gives  xy=9.7,  and  from 
these  last  two  equations,  x=9  or  3,  and  y=3  or  9. 

There  are  a  great  variety  of  circumstances  that  may  come 
in  aid,  or  deter  the  solution  of  equations  ;  but  it  is  not  proper 
to  notice  them  in  an  elementary  work  like  this.  For  a  more 
full  development  of  these  particulars  in  equations,  see  Robin- 
son's Algebra,  University  Edition. 

(Art.  103.)  It  is  not  essential  that  the  unknown  quantity 
should  be  in  involved  literally  to  its  first  and  second  powers  ; 
it  is  only  essential  that  the  index  of  one  power  should  be 
double  that  of  the  other.*  In  such  cases,  the  equations  can  be 
resolved  as  quadratics.  For  example,  x^ — 4x^=z621  is  an 
impure  equation  of  the  sixth  degree,  yet  with  a  view  to  its 
solution,  it  may  be  called  a  quadratic.     For  we  can  assume 


*  From  this  and  the  following  article  we  perceive  that  the  term  quad- 
ratic equations,  is  far  more  proper  and  comprehensive  than  equations  of 
the  second  degree. 

"We  speak  of  this  because  it  has  been  suggested  to  us,  that  the  modern 
rules  of  science  required  the  systematic  use  of  the  term  equations  of 
the  first,  second,  third,  &c.,  degrees.  The  author  of  this  work  is  modern 
in  all  his  views,  and  is  an  advocate  for  modern  improvements ;  but  it 
must  be  improvements,  not  merely  varieties,  or  changes  in  technicalities. 


19S  ELEMENTARY  ALGEBRA. 

y=si?;  then  y^=x^,  and  the  equation  becomes  i/^ — 43/=621, 
a  quadratic  in  relation  to  y,  giving  y='21,  or  — 23. 

Therefore,  .'    .     .     ar'=27  or       —23 

And       ....       ar=3     or   V— 23 

There  are  other  values  of  x;  but  it  would  be  improper  to 
seek  for  them  now;  such  inquiries  belong  to  the  higher  order 
of  equations.* 

For  another  example,  take  a? — x^=56,  to  find  the  values 
of  X. 

Here  we  perceive  one  exponent  of  x  is  double  that  of  the 
other;  it  is  therefore  essentially  a  quadratic. 

Such  cases  can  be  made  clear  by  assuming  the  lowest 
power  of  the  unkno'svn  quantity  equal  to  any  single  letter. 
In  the  present  case,  assume  y=x'^ ;  then  y^=x^,  and  the  equa- 
tion becomes    ,     .     .     y^ — y=56 

A  solution  gives  y=8,  or  — 7,  and  by  returning  to  the 

3.  3.  i 

assumption,  y=x^,  we  find  X"  =  S,  or  x-=2,  or  x=4. 

(Art.  104.)  When  a  compound  quantity  appears  under 
different  powers  or  fractional  exponents,  one  exponent  being 
double  of  the  other,  we  may  put  the  quantity  equal  to  a  single 
letter,  and  make  its  quadratic  form  apparent  and  simple.  For 
example,  suppose  the  values  of  x  were  required  in  the  equation 
2x''-]-3x+9—5j2x^-\-3x-\-9=G 

Assume J2x^-jr^x-i-9=y 

Then  by  involution, .     .        2x^ -\-3x-{-9  =y-         (A) 
And  the  equation  becomes       .    y^ — 5y=6  (jB) 

Which  equation  gives  y=6  or  — 1.  These  values  of  y, 
substituted  for  y  in  equation  (A),  give 

2r^-l- 3a; -1-9=36 

Or, 2ar»-f3a;-|-9=l 

From  the  first  of  these  we  find    .     .     x=3  or  — 4^. 

*  See  Algebra,  University  Edition. 


EQUATIONS.  199 


From  the  last,  we  find  x=l( — 3±:J — 55),  imaginary 
quantities. 

We  give  a  few  examples  to  fix  the  principles  explained  in 
Articles  103  and  104. 

1.  Given  x-\-3-\-2(x-{-3)'^=35,  to  find  one  value  of  x. 

Ans.  x=22. 

2.  Given  (f+2yy-\-4(i/^+2y)=96,  to  find  one  value  of  y. 

Ans.  y=2. 

3.  Given  10+a; — (10+a;)2  =  12,  to  find  one  value  of  x. 

Ans.  x=Q, 

4.  Given  (  --Hy  J+  [  -+y  )  =30  to  find  y. 

Ans.  y=3  or  2,  or  — 3±73. 

5.  Given  (a;-l-12)^+(a:-i-12)^=6,  to  find  the  values  of  x. 

Ans.  x—4  or  69. 

6.  Given  (a;4-«)^  +  25(a;+a)*=35^  to  find  the  values  of  ar. 

Ans.  ir=Z»^ — a  or  81^'* — a. 
It  is  very  seldom  that  problems  produce  such  compound 
equations  as  the  last  six,  or  indeed  never  will  unless  expressly 
designed  so  to  do.     The  following  is  one : 

1.  A  poulterer  going  to  market  to  buy  turkeys,  met  with 
four  flocks.  In  the  second,  were  6  more  than  3  times  the 
square  root  of  double  the  number  in  the  first.  The  third 
contained  3  times  as  many  as  in  the  first  and  second ;  and  the 
fourth  contained  6  more  than  the  square  of  one-third  the  num- 
ber in  the  third;  and  the  Avhole  number  was  1938.  How 
many  were  in  each  flock?  Ans.  18,  24,  126,  1770. 

Let  ....     2x^=  the  number  in  the  first. 
Then     .     .     6ar-|-6=  the  number  in  the  second, 
3(2x^-\-6x-\-6)—  the  number  in  the  third, 
(2x^-\-6x-\-6y-{-6=  the  number  in  the  fourth. 
Assume  2x^-\-6x-\-6==y.     Then  the  whole  sum  is 
y^-|-4y+6=1938 


200  ELEMENTARY  ALGEBRA. 

Subtracting  2  from  both  members,  and  extracting  square 
root,  we  have      .     .     .    2/-|-2=44 

We  do  not  take  the  minus  sign,  for  minus  cannot  apply  to 
this  problem. 

From  the  assumed  equation,  we  have 

2.  If  a  certain  number  be  increased  by  3,  and  the  square 
root  of  the  sum  taken  and  added  to  the  number,  the  sum  will 
be  17.     What  is  the  number  ?  Ans.  13. 

3.  The  square  of  a  certain  number,  and  1 1  times  the  num- 
ber makes  80.    What  is  the  number  ?  Ans.  5. 

4.  Find  two  numbers,  sucli  that  the  less  may  be  to  the 
greater  as  the  gi-eater  is  to  12,  and  that  the  sum  of  their 
squares  may  be  45.  Ans.  3  and  6. 

6.  "What  two  numbers  are  those,  whose  difference  is  3,  and 
the  difference  of  their  cubes  189  ?  Ans.  3  and  6. 

6.  What  two  numbers  are  those,  whose  sum  is  5,  and  the 
sum  of  their  cubes  35  ?  Ans.  2  and  3. 

7.  A  merchant  has  a  piece  of  broadcloth  and  a  piece  of 
eilk.  The  number  of  yards  in  both  is  110;  and  if  the  square 
of  the  number  of  yards  of  silk  be  subtracted  from  80  times 
the  number  of  yards  of  broadcloth,  the  difference  will  be  400, 
How  many  yards  are  there  in  each  piece  ? 

Ans.  60  of  silk;  50  of  broadcloth. 

8.  A  is  4  years  older  than  B ;  and  the  sum  of  the  squares 
of  their  ages  is  976.     What  are  their  ages  ? 

Ans.  A's  age,  24  years;  B's,  20  years. 

9.  Divide  the  number  10  into  two  such  parts,  that  the 
square  of  4  times  the  less  part,  may  be  112  more  than  the 
square  of  2  times  the  greater.  Ans.  4  and  6. 

10.  Find  two  numbers,  such  that  the  sum  of  their  squares 
may  be  89,  and  their  sum  multiplied  by  the  greater,  may 
produce  104.  A7is.  5  and  8 


EQUATIONS.  201 

11.  What  number  is  that,  which,  being  divided  by  the 
product  of  its  two  digits,  the  quotient  is  5];  but  when  9  is 
subtracted  from  it,  there  remains  a  number  having  the  same 
digits  inverted  ?  Ans.  32. 

12.  Divide  20  into  three  parts,  such  that  the  continual 
product  of  all  three  may  be  270,  and  that  the  difference  of 
the  first  and  second  may  be  2  less  than  the  difference  of  the 
second  and  third.  Ans.  5,  6,  and  9. 

13.  A  regiment  of  soldiers,  consisting  of  1066,  formed  into 
two  squares,  one  of  which  has  four  men  more  in  a  side  than 
the  other.  What  number  of  men  are  in  a  side  of  each  of  the 
squares?  Ans.  21  and  25. 

14.  The  plate  of  a  lookingglass  is  18  inches  by  12,  and  is 
to  be  framed  with  a  frame  of  equal  width,  whose  area  is  to 
be  equal  to  that  of  the  glass.  Required  the  width  of  the 
frame.  Ans.  3  inches. 

15.  A  square  courtyard  has  a  rectangular  gravel  walk 
round  it.  The  side  of  the  court  wants  two  yards  of  being- 
six  times  the  width  of  the  gravel  walk,  and  the  number  of 
square  yards  in  the  walk  exceeds  the  number  of  yards  in  the 
periphery  of  the  court  by  164.  Required  the  area  of  the 
court.  Ans.  256  yards. 

16.  A  and  B  start  at  the  same  time  to  travel  150  miles;  A 
travels  3  miles  an  hour  faster  than  B,  and  finishes  his  jour- 
ney 8i  hours  before  him ;  at  what  rate  per  hour  did  each 
travel  ?  Ans.  9  and  6  miles  per  hour. 

17.  A  company  at  a  tavern  had  1  dollar  and  75  cents  to 
pay ;  but  before  the  bill  was  paid  two  of  them  went  away, 
when  those  who  remained  had  each  10  cents  more  to  pay; 
how  many  were  in  the  company  at  first  ?  Ans.  7. 

18.  A  set  out  from  C,  toward  D,  and  traveled  7  miles  a 
day.  After  he  had  gone  32  miles,  B  set  out  from  D  toward 
C,  and  went  every  day  j\  of  the  whole  journey;  and  after  he 


202  ELEMENTARY  ALGEBRA. 

had  traveled  as  many  days  as  he  went  miles  in  a  day,  he  met 
A.     Required  the  distance  from  C  to  D. 

Ans.  76  or  152  miles;  both  numbers  will  answer  the  con- 
dition. 

19.  A  farmer  received  24  dollars  for  a  certain  quantity  of 
wheat,  and  an  equal  sum  at  a  price  25  cents  less  by  the 
bushel  for  a  quantity  of  barley,  which  exceeded  the  quantity 
of  wheat  by  16  bushels.  How  many  bushels  were  there  of 
each  ?  Ans.  32  bushels  of  wheat,  and  48  of  barley. 

20.  A  laborer  dug  two  trenches,  one  of  which  was  6  yards 
longer  than  the  other,  for  17  pounds,  16  shillings,  and  the 
digging  of  each  of  them  cost  as  many  shillings  per  yard  as 
there  were  yards  in  its  length.     What  was  the  length  of  each  ? 

Ans.   10  and  16  yards. 

21.  A  and  B  set  out  from  two  towns  which  were  distant 
from  each  other  247  miles,  and  traveled  the  direct  road  till 
they  met.  A  went  9  miles  a  day,  and  the  number  of  days  at 
the  end  of  which  they  met,  was  greater,  by  3,  than  the  num- 
ber of  miles  which  B  went  in  a  day.  How  many  miles  did 
each  travel?  Ans.  A,  117,  and  B  130  miles. 

22.  The  fore  wheels  of  a  carriage  make  6  revolutions  more 
than  the  hind  wheels,  in  going  120  yards ;  but  if  the  circum- 
ference of  each  wheel  be  increased  1  yard,  they  will  make  only 
4  revolutions  more  than  the  hind  wheels,  in  the  same  distance  ; 
required  the  circumference  of  each  wheel. 

Ans.  4  and  5  yards. 

23.  There  are  two  numbers  whose  product  is  120.  If  2 
be  added  to  the  lesser,  and  3  subtracted  from  the  greater,  the 
product  of  the  sum  and  remainder  will  also  be  120.  What 
are  the  numbers  ?  Ans.  15  and  8. 

24.  There  are  two  numbers,  the  sum  of  whose  squares 
exceeds  twice  their  product,  by  4,  and  the  difference  of  tlieir 
squares  exceeds  half  their  product,  by  4;  required  the 
numbers.  A7is.  6  and  8. 


EQUATIONS.  203 

25.  "What  two  numbers  are  those,  which  being  both  mul- 
tiplied by  27,  the  first  product  is  a  square,  and  the  second 
the  root  of  that  square ;  but  being  both  multiplied  by  3,  the 
first  product  is  a  cube,  and  the  second  the  root  of  that  cube  ? 

Ans.  243  and  3. 

26.  A  man  bought  a  horse,  which  he  sold,  after  some  time, 
for  24  dollars.  At  this  sale  he  loses  as  much  per  cent,  upon 
the  price  of  his  purchase  as  the  horse  cost  him.  What  did 
he  pay  for  the  horse  ? 

Ans.  He  paid  f  60  or  $40 ;  the  problem  does  not  decide 
which  sum. 

27.  What  two  numbers  are  those  whose  product  is  equal  to 
the  difference  of  theii  squares ;  and  the  greater  number  is  to 
the  less  as  3  to  2  ?  Ans.  No  such  numbers  exist. 

28.  What  two  numbers  are  those,  the  double  of  whose 
product  is  less  than  the  sum  of  their  squares  by  9,  and  half 
their  product  is  less  than  the  difference  of  their  squares  by  9  ? 

Ans,  The  numbers  are  9  and  12. 
Will  the  student  show  that  examples  24  and  28  are  essen- 
tially the  same. 


SECTION   V. 


ARITHMETICAL    PROGRESSION. 

(Art.  105.)  A  series  of  numbers  or  quantities,  increasing 
or  decreasing  by  the  same  diflference,  from  term  to  term,  is 
called  arithmetical  progression. 

Thus,  2,  4,  6,  8,  10,  12,  &c.,  is  an  increasing  or  ascending 
arithmetical  series,  having  a  common  diflference  of  2;  and  20, 
17,  14,  11,  8,  &c.,  is  a  decreasing  series,  whose  common  dif- 
ference is  3. 

"We  can  more  readily  investigate  the  properties  of  an  arith- 
metical series  from  literal  than  from  numeral  terms.  Thus, 
let  a  represent  the  first  term  of  a  series,  and  d  the  common 
diflference.     Then 

a,  (a-^d),  (a-\-2d),  (a-{-3d),  (a-{-4d),  &c., 
represents  an  ascending  series ;  and 

a,  (a-^),  (a— 2d),  (a— 3d),  (a—id),  <fec., 
represents  a  descending  series. 

Observe  that  the  coeflScient  of  d  in  any  term,  is  equal  to 
the  number  of  the  preceding  term. 

The  first  term  exists  without  the  common  diflference.  All 
other  terms  consist  of  the  first  term  and  the  common  diflfer* 
ence  multiplied  by  one  less  than  the  number  of  terms. 


ARITHMETICAL  PROGRESSION.  205 

Thus,  if  the  first  term  of  an  arithmetical  series  is  a,  and  d 
the  common  difference,  the  tenth  term  would  be  expressed  by 

The  1 7th  term  by  .  a-\-\&d 
The  53d  term  by  .  a-\-52d 
The  wth    term  by       .     a-\-{n — \)d 

When  the  series  is  decreasing,  the  sign  to  the  term  contain- 
ing d  will  be  minus,  the  20th  term,  for  example,  would  be 

a— 19g? 
The  wth  term     .     .     .     a — {n — 1  )d 

We  add  a  few  examples  to  exercise  the  pupil  in  finding  any 
term  of  a  series,  when  the  first  term,  a,  and  the  common  dif- 
ference, c?,  are  given. 

1.  Whena=2    and  c?=3,    what  is  the  10th    term? 

Ans.  29. 

2.  When  a= 3    and  fl?=2,    what  is  the  12th    term? 

Ans.  25. 

3.  When  a=7    and  d=\0,  what  is  the  21st     term  ? 

Ans.  207. 

4.  When  a=l    and  d—\y    what  is  the  100th  term  ? 

Ans.  50|, 
6.  When  a=3    and  <f=i,    what  is  the  100th  term  ? 

Ans.     36. 

6.  When  a=0    and  c?=|,    what  is  the  89th    term  ? 

Ans.     1 1 . 

7.  When  a=Q    and  c^=— i,  what  is  the  20th  term  ? 

Ans.  —31. 

8.  When  a=30  and  c?= — 3,  what  is  the  31st  term  ? 

Ans.  — 60. 

Wherever  the  series  is  supposed  to  terminate,  is  the  last 
term,  and  if  such  term  be  designated  by  Z,  and  the  number 
of  tenns  by  n,  the  last  term    must   bfe   a+(rj — \)d,  or 


ELEMENTARY  ALGEBRA. 

a — [n — '\)d,  according  as  the  series  may  be  ascending  or 
descending,  which  we  draw  from  inspection. 

Hence,  ....     L=adz(n — l)d        (A) 

(Art.  106.)  It  is  manifest,  that  the  sum  of  the  terms  will 
be  the  same,  in  whatever  order  they  are  written. 

Take,  for  instance,  the  series  .  3,  5,  7,  9,11, 
And  the  same  inverted,  .  .  .11,  9,  7,  5,  3. 
The  sums  of  the  terms  will  be  14,  14,  14,  14,  14. 
Take       .  .a  a-\-  d,    a-]-2d,    a-\-2d,    a-\-4d, 

Inverted,     .     .   a-\-4d,    a-\-3d,    a-\-2d,    a-\-  d,    a 
Sums,     .     .       2a+4f^,  2a-{-4d,  2a-\-4d,  2a-{-4d,  2a-\-4d, 

Here  we  discover  the  important  property,  that,  in  arithme- 
tical progression,  the  sum  of  the  extremes  is  equal  to  the  sum  of 
any  other  two  terms  equally  distant  from  the  extremes.  Also, 
that  twice  the  sum  of  any  series  is  equal  to  the  extremes,  or  first 
and  last  term  repeated  as  many  times  as  the  series  contains  terms. 

Hence,  if  S  represents  the  sum  of  a  series,  and  n  the  num- 
ber of  terms,  a  the  first  term,  and  L  the  last  term,  we  shall 
have 2S=n{a-^L) 

Or, S^lia-^L)         {B) 

The  two  equations  (.^4)  and  [B)  contain  five  quantities,  a, 
d,  L,  %,  and  S;  any  three  of  them  being  given,  the  other  two 
can  be  determined. 

Two  independent  equations  are  sufficient  to  determine  two 
unknown  quantities  (Art.  45),  and  it  is  immaterial  which  two 
are  unknown,  if  the  other  three  are  given. 

By  examining  the  two  equations  they  will  become  familiar. 
X=a+(«-l)/        {A) 


ARITHMETICAL  PROGRESSION.  207 

Equation  (A)  and  (B)  furnish  all  the  rules  given  in  Arith- 
metics in  relation  to  arithmetical  progression. 

For  instance,  the  rule  to  find  the  last  term  of  any  arithme- 
tical series,  is  equation  (A)  put  in  words,  thus : 

Rule  . — Multiply  the  common  difference  by  the  nuniber  of 
terms  less  one,  and  to  the  product  add  the  first  term. 

A  rule  for  finding  the  sum  of  any  series,  we  draw  from 
equation  (^),  thus : 

Rule  . — Multiply  the  sum  of  the  extremes  hy  half  the  number 
of  terms. 

EXAMPLES. 

1.  The  first  term  of  an  arithmetical  series  is  5,  the  last 
term  92,  and  the  number  of  terms  30.  What  is  the  sum  of 
the  terms?  Ans.  1465. 

2.  The  first  term  of  an  arithmetical  series  is  2,  the  number 
of  terms  10,  and  the  last  term  30.  What  is  the  sum  of  the 
terms?  Ans.  160. 

3.  The  first  term  of  an  arithmetical  series  is  5,  the  common 
difierence  3,  and  the  number  of  terms  30.  What  is  the  last 
term?  Ans.  92. 

4.  The  first  term  of  an  arithmetical  series  is  7,  the  last 
term  207,  and  the  number  of  terms  21.  What  is  the  sum  of 
the  terms  ?  Ans.  2247. 

5.  The  first  term  of  an  arithmetical  series  is  6,  the  last 
term  — 3^,  and  the  number  of  terms  20.  What  is  the  sum 
of  the  terms  ?  Ans.  25. 

The  two  equations  [A)  and  [B)  cover  the  whole  subject 
of  arithmetical  progression,  when  any  three  of  the  five  quan- 
tities are  given ;  for  there  would  be  two  unknown  quantities, 
and  we  have  two  equations,  which  are  sufficient  to  find  them ; 
we,  therefore,  give  the  following  miscellaneous  examples  : 

Use  the  equations  vMhout  modification  or  change,  by  putting 
in  the  given  values  just  as  they  stand,  and  afterward  reduce 
them  as  numeral  equations. 


208  ELEMENTARY   ALGEBRA. 

EXAMPLES. 

1.  The  sum  of  an  arithmetical  series  is  1455,  the  first  term 
5,  and  the  number  of  terms  30.  What  is  the  common 
difference?  A7is.  3. 

Here,      ^=1455,  a=5,  n=SO.    Z  and  cf  are  sought. 
Equation  (B),  1455=(5+Z)15.     Reduced,  Z=92. 
Equation  (A),       92=5-\-29d.         Reduced,  d=S,  Am. 

2.  The  sum  of  an  arithmetical  series  is  567,  the  first  term 
7,  and  the  common  difference  2.  What  is  the  number  of 
terms?  Ans.  21. 

Here,      >S'=567,  a=7,  d=2.    L  and  n  are  sought. 
Equation  {A),      Z=7-\-2n — 2=5+ 2w 

Equation  (J5),   667=(7+5-h2w)|=6w+»=» 

Or, n^+6n+9=576 

w+3=24,  or  «=21,  Ans. 

3.  Find  seven  arithmetical  means  between  1  and  49. 
Observe  that  the  series  must  consist  of  9  terms. 
Hence,  a=l,  Z=49,  n=9. 

Ans.  7,  13,   19,  25,  31,  37,  43. 

4.  The  first  term  of  an  arithmetical  series  is  1,  the  sum  of 
the  terms  280,  the  number  of  terms  32.  What  is  the  com- 
mon difference,  and  the  last  term  ?  Ans.  d=^,  Z=16^. 

5.  Insert  three  arithmetical  means  between  i  and  ^. 

Ans.  The  means  are  f ,  /^,  |{. 

6.  Insert  five  arithmetical  means  between  5  and  1 5. 

Ans.  The  means  ars  6|,  81,   10,   llf,   13^. 

7.  Suppose  100  balls  be  placed  in  a  straight  line,  at  the 
distance  of  a  yard  from  each  other;  how  far  must  a  person 
travel  to  bring  them  one  by  one  to  a  box  placed  at  the  distance 
of  a  yard  from  the  first  ball  ? 

Ans.  6  miles  and  1300  yards. 


ARITHMETICAL  PROGRESSION.  209 

8.  A  speculator  bought  47  house  lots  in  a  certain  village, 
giving  10  dollars  for  the  first,  30  dollars  for  the  second,  50 
dollars  for  the  third,  and  so  on.  What  did  he  pay  for  the 
whole  47  ?  Ans.  $22090. 

9.  In  gathering  up  a,  certain  number  of  balls,  placed  on  the 
ground  in  a  straight  line,  at  the  distance  of  2  yards  from  each 
other,  the  first  being  placed  2  yards  from  the  box  in  which 
they  were  deposited,  a  man,  starting  from  the  box,  traveled 
11  miles  and  840  yards.     How  many  balls  were  there  ? 

Ans.   100. 

10.  How  many  strokes  do  the  clocks  of  Venice,  which  go 
on  to  24  o'clock,  strike  in  a  day  ?  Ans.  300. 

11.  In  a  descending  arithmetical  series  the  first  term  is  730, 
the  common  difi'erence  2,  and  the  last  term  2.  What  is  the 
number  of  terms  ?  Ans.  365. 

12.  The  sum  of  the  terms  of  an  arithmetical  series  is  280, 
the  first  term  1 ,  and  the  number  of  terms  32.  What  is  the 
common  difference  ?  Ans.  -J. 

13.  The  sum  of  the  terms  of  an  arithmetical  series  is  950, 
thie  common  difference  3,  and  the  number  of  terms  25.  What 
is  the  first  term  ?  Ans.  2. 

14.  What  is  the  sum  of  n  terms  of  the  series  1 ,  2,  3,  4,  5, 

&C.1  Ans.  S=:-(l-{-n) 

2^  / 

15.  Suppose  a  man  owes  1000  dollars,  what  sum  shall  he 
pay  daily  so  as  to  cancel  the  debt,  principal  and  interest,  at 
the  end  of  a  year,  reckoning  it  at  6  per  cent,  simple  interest  ? 

Divide  1000  dollars  by  365,  and  call  the  quotient  a.  This 
would  be  the  sum  he  must  pay  daily,  provided  there  were  no 
interest  to  be  paid. 

Cast  the  interest  on  a  for  one  day,  at  6  per  cent,  and  call 
this  interest  i. 
18 


210  ELEMENTARY  ALGEBRA. 

Then  the  first  day  he  must  pay  a-\-i 
The  second  day,      ....     a-f-Si 
The  third  day, a-{-3i;  and  so,  on  in  arith- 
metical progression. 

The  last  day  he  must  pay      .     a-{-365i 


Altogether,  he  must  pay  .     I I  c 


365. 


Or,  he  must  pay  daily,      .     .     a4-183i=  the  answer. 

(Art.  107.)  Bodies  falling  near  the  surface  of  the  earth,  sind 
unresisted  by  the  atmosphere,  fall  in  the  first  second  of  time 
16Jj  feet,  and  increase  the  distance  which  they  fall  2(16^^) 
feet  every  second.  Hence,  lGj\  feet  may  be  considered  the 
first  term  of  an  arithmetical  series,  and  2(\6j\)  the  common 
difference.  We  call  IGJ^-  feet  ff,  the  symbol  for  gravity. 
Then  ff  is  the  first  term  of  an  arithmetical  series,  and  2g  the 
common  difi*erence.  Hence,  g,  3g,  bg,  Ig^  9^,  &c.,  are  the 
spaces  corresponding  to  1,  2,  3,  4,  &c.,  seconds. 

These  facts  being  admitted,  show  a  formula  for  the  fall  of 
a  body  in  10  seconds,  and  for  its  fall  the  last  second  of  the 
ten, 

From  (A)     .     .     .     L^g-^-^'Stg 

From(J5)     .     .     .     S=b(2g-{-\^g) 

Hence,  its  fall  during  the  last  second  of  the  ten  is  19^,  and 
the  whole  space  fallen  through  is  100^,  which  is  the  square  of 
the  seconds  multiplied  by  the  force  of  gravity,  and  this  is  the 
general  rule  in  Astronomy. 

But  this  manner  of  arriving  at  the  result  is  not  recommended, 
except  as  an  exercise  in  progression. 

PROBLEMS   IN    ARITHMETICAL   PROGRESSION 

TO   WHICH   THE    FRKCEDINO    FORMULAS,  {A)   AND    (B), 
DO   NOT   IMMEDIATELY    APPLY. 

(Art.  108.)  When  three  quantities  are  in  arithmetical  pro- 
gression, it  is  evident  that  the  middle  one  must  be  the  exact 
inean  of  the  three,  otherwise,  it  would  not  be  arithmetical 


ARITHMETICAL  PROGRESSION.  211 

progression ;  therefore  the  sum  of  the  extremes  must  be  double 
that  of  the  mean. 

Take,  for  example,  any  three  consecutive  terms  of  a  series, 

as a-i-2c?,     a-\-^d,     a-\-4d 

and  we  perceive,  by  inspection,  that  the  sum  of  the  extremes 
is  double  that  of  the  mean. 

When  there  are  four  terms,  the  sum  of  the  extremes  is 
equal  to  the  sum  of  the  means,  by  (Art.  106). 

To  facilitate  the  solution  of  problems,  when  three  terms  are 
in  question,  let  them  be  represented  by  (x — y),  x,  (x-{-y), 
y  being  the  common  difference. 

When  four  numbers  are  in  question,  let  them  be  repre- 
sented by  (a:— 3y),  {x—y),  {xArv),  {x^^y)y  ^y  being 
the  common  difference. 

So  in  general  for  any  other  number,  assume  such  terms 
that  th^  common  difference  will  disappear  by  addition, 

EXAMPLES. 

1.  Three  numbers  are  in  arithmetical  progression,  the 
product  of  the  first  and  second  is  15,  and  of  the  first  and  third 
is  21.     What  are  the  numbers  ?  Ans.  3,  5,  and  7. 

2.  There  are  four  numbers  in  arithmetical  progression,  the 
sum  of  the  two  means  is  25,  and  the  second,  multiplied  by 
the  common  difference  is  50.     What  are  the  numbers  ? 

Ans.  5,  10,  15,  and  20. 

3.  There  are  four  numbers  in  arithmetical  progression,  the 
product  of  the  first  and  third  is  5,  and  of  the  second  and 
fourth  is  21.     What  are  the  numbers  1    Ans.  1,  3,  5,  and  7. 

4.  There  are  five  numbers  in  arithmetical  progression,  the 
sum  of  these  numbers  is  65,  and  the  sum  of  their  squares 
1005.     What  are  the  numbers?    Ans.  5,  9,  13,  17,  and  21. 

Let  x^=  the  middle  term,  arid  y  the  common  difference. 


212  ELEMENTARY  ALGEBRA. 

Then  x — 2y,  x — y,  x,  x-\-y,  x-\-2y,  will  represent  the 
numbers,  and  their  sum  will  be  5x=65,  or  a;=13.  Also,  the 
sum  of  their  squares  will  be 

5ar^+10y2=l005,  or  a^^-f  23/2= 201. 

5.  The  sum  of  three  numbers  in  arithmetical  progression 
is  15,  and  their  continued  product  is  105.  What  are  the 
numbers?  "  Ans.  3,  5,  and  7. 

6.  There  are  three  numbers  in  arithmetical  progression, 
their  sum  is  18,  and  the  sum  of  their  squares  158.  What 
are  those  numbers  ?  '  Ans.  1,  6,  and  1 1. 

7.  Find  three  numbers  in  arithmetical  progression,  such 
that  the  sum  of  their  squares  may  be  56,  and  the  sum  arising 
by  adding  together  once  the  first  and  twice  the  second,  and 
thrice  the  third,  may  amount  to  28.  Aiis.  2,  4,  6. 

8.  Find  three  numbers  having  equal  differences,  so  that 
their  sum  may  be  12,  and  the  sum  of  their  fourth  powers 
962.  Ans.  3,  4,  5. 

9.  Find  three  numbers  having  equal  differences,  and  such 
that  the  square  of  the  least  added  to  the  product  of  the  two 
greater,  may  make  28,  but  the  square  of  the  greatest  added 
to  the  product  of  the  two  less,  may  make  44.      Ans.  2,  4,  6. 

10.  Find  three  numbers  in  arithmetical  progression,  such 
that  their  sum  shall  be  1 5,  and  the  sum  of  their  squares  93. 

Ans.  2,  5,  and  8. 

11.  Find  three  numbers  in  arithmetical  progression,  such 
that  the  sum  of  the  first  and  third  shall  be  8,  and  the  sum  of 
the  squares  of  the  second  and  third  shall  be  52. 

Ans.  2,  4,  and  6. 

12.  Find  four  numbers  in  arithmetical  progression,  such 
that  the  sura  of  the  first  and  fourth  shall  be  13,  and  the  dif- 
ference of  the  squares  of  the  two  means  sliall  be  39. 

Ans.  2,  5,  8,  and  11. 

13.  Find  seven  numbers  in  arithmetical  progression,  such 


ARITHMETICAL  PROGRESSION. 

that  the  sum  of  the  first  and  sixth  shall  he  14,  and  the  pro- 
duct of  the  third  and  fifth  shall  be  60. 

Ans.  2,  4,  6,  8,  10,  12,  and  14. 

15.  Find  five  numbers  in  arithmetical  progression,  such 
that  their  sum  shall  be  25,  and  their  continued  product  945. 

Ans.  1,  3,  5,  7,  and  9. 

16.  Find  four  numbers  in  arithmetical  progression,  such 
that  the  dijQference  of  the  squares  of  the  first  and  second  shall 
be  12,  and  the  diflference  of  the  squares  of  the  third  and  fourth 
shall  be  28.  Ans.  2,  4,  6,  and  8. 


GEOMETRICAL  PROGRESSION. 

(Art.  109.)  When  a  series  of  numbers  or  quantities  in- 
crease or  decrease  by  a  constant  multiplier  from  term  to  term, 
the  numbers  or  quantities  are  said  to  be  in  geometrical  pro- 
ffression,  and  the  constant  multiplier  is  called  the  ratio. 

Thus,  let  a  be  the  first  term  of  the  progression,  and  r  the 
ratio,  then  a,  ar,  ar^,  ar^,  ar*,  &c.,  will  represent  the  series. 

If  r  is  greater  than  1 ,  the  series  will  be  ascending;  if  less 
than  1 ,  the  series  will  be  descending,  and  if  r=  1 ,  every  terra 
of  the  series  will  be  the  same  in  value. 

For  example,  2,  6,  18,  54, 162,  &c.,  is  a  geometrical  series 
in  which  the  first  term  a  is  2,  and  the  ratio  is  3. 

The  series  9,  3,  1,  i,  i,  ^\,  <fec.,  is  also  a  geometrical  series 
in  which  the  first  term  a  is  9,  and  the  multiplier,  the  ratio. 

The  series  3,  3,  3,  3,  <fec.,  is  also  a  geometrical  series  in 
which  the  first  term  a,  is  3,  and  the  multiplier,  the  ratio,  is  1. 

Geometry  compares  magnitudes,  and  inquires  how  many 
times  one  magnitude  is  greater  than  another,  and  thus,  in  the 


814  ELEMENTARY    ALGEBRA. 

series,  2,  4,  8,  16,  <fec.,  4  is  two  times  2,  8  is  2  times  4,  &,c.- 
hence,  numbers  so  compared,  and  a  regular  series  thus 
obtained,  is  called  a  geometrical  series. 

(Art.  110.)  In  any  given  series  we  may  find  the  ratio,  by 
dividing  any  term  by  its  preceding  term. 

(Art.  111.)  Taking  the  general  series  a,  ar,  ar^y  ar^,  ar"^, 
&c.,  under  inspection,  we  find  that  the^r*^  power  of  r  is  a 
factor  in  the  second  term,  the  second  power  of  r  in  the  third 
term,  the  third  power  of  r  in  the  fourth  term,  and  thus,  uni- 
versally, the  power  of  the  ratio  in  any  term,  is  one  less  than 
the  number  of  the  term. 

The  first  term  is  a  factor  in  every  term.  Hence,  the  10th 
term  of  this  general  series  is  ar^.  The  1 7th  term  would  be 
ar^^.  The  25th  term  would  be  ar^S  and,  in  general,  the  wth 
term  would  be  ar'^~\ 

Therefore,  if  n  represent  the  number  of  terms  in  any 
series,  and  L  the  last  term,  then 

Z=ar™-»         (1) 

Wherever  a  series  is  supposed  to  terminate,  is  the  last  term, 
and  equation  ( 1 )  is  a  general  representation  of  it ;  and  if  we 
multiply  that  equation  by  r,  we  shall  have 
rL=ar^ 

(Art.  112.)  Let  S  represent  the  sum  of  any  geometrical 
series,  then  we  have 

S=s:a-\-a.r'\-ar^'^ai*,  &c.,  to  ar*"^ 

Multiply  this  equation  by  r,  and  we  have 

rSs=ar-\-ar'-\-a7^,  <fec.,  to  ar"~'-}-ar" 

Subtracting  the  upper  equation  from  the  lower,  and  observ- 
ing that rL=ar^ 

Then     .     .     .     (r--l)S=rL—a 

Therefore,      .     .     .      5« (2) 


GEOMETRICAL  PROGRESSION.  215 

As  the  equations  (1)  and  (2)  are  fundamental,  and  cover 
the  whole  subject  of  geometrical  progression,  let  them  be 
brought  together  for  critical  inspection. 

Here  we  perceive  five  quantities,  a,  r,  n,  Z,  and  S^  and  any 
three  of  them  being  given  in  any  problem,  the  other  two  can 
be  determined  from  the  equations, 

L^ar-^-"-         (1) 
8="^       (2) 

"These  two  equations  furnish  the  rules  given  for  the  opera- 
tions in  common  arithmetic. 

Thus,  in  almost  every  Arithmetic,  the  rules  for  finding  the 
ifest  term  of  any  arithmetical  series  is  expressed  in  the  follow- 
ing words : 

R  tj  L  E . — Raise  the  ratio  to  a  power  one  less  than  the  number 
of  terms,  and  multiply  that  number  by  the  first  term. 

This  rule  is  simply  equation  ( 1 )  put  in  words. 

Equation  (2)  gives  the  following  rule  for  the  sum  of  a 
series. 

Rule  . — Multiply  the  last  term  by  the  ratio,  and  from  the 
product  subtract  the  first  term,  and  divide  the  remairider  by  the 
ratio  less  one. 

GENERAL    EXAMPLES    IN    GEOMETRICAL 
PROGRESSION. 

1.  What  is  the  ratio  of  the  series  2,  6,  18,  54,  &c.? 

Ans.  3. 

2.  What  is  the  ratio  of  the  series  5,  20,  80,  &c.?  Ans.  4. 

3.  What  is  the  ratio  of  the  series  |-,  |-,  2-\,  &c.l    Ans.  ?j, 

4.  What  is  the  ratio  of  the  series  y^.^  ^i_^  toV o»  ^^• 

Ans.  yV- 

5.  What  is  the  ratio  of  the  series  f,  1,  ^,  &c.?      Ans.     |. 


216  ELEMENTARY  ALGEBRA. 

6.  What  is  the  ratio  of  the  series  8,  20,  60,  &c.?  Ans.  ^ 

7.  What  is  the  ratio  of  the  series  -,  — ,  — ,  &c.?    Ans.    — 

8.  What  is  the  ratio  of  the  series  a,  — h,  -\ —  &c.? 

Ans. • 

a 

9.  What  is  the  11th  term  of  the  series  1,  2,  4,  &c.? 

Ans.  1024. 

10.  What  is  the    9th  term  of  the  series  5,  20,  80,  &c.? 

Ans.  327680. 

11.  What  is  the    8th  term  of  the  series  2,  6,  18,  &c.? 

Ans.  4374. 

12.  What  is  the    6th  term  of  the  series  1,  |,f^,(fec. 

Ans.  /^V*- 

13.  What  is  the  sum  of  8  terms  of  the  series  2,  6,  18,  &c.? 

Ans.  6560. 

,^.  „    rZ—a    3*4374—2     ^^^^ 

(2)         S= -= =6560 

^   ^  r— 1  2 

14.  What  is  the  sum  of  10  terms  of  the  series  4,  12,  36, 
&c.?  Ans,  118096. 

15.  What  is  the  sum  of  9  terms  of  the  series  5,  20,  80, 
&c.?  Ans.  436905. 

16.  What  is  the  sum  of  5  terms  of  the  series  3,  4|-,  63, 
&c.?  ^  Ans.  39y9^. 

17.  What  is  the  sum  of  10  terms  of  the  series  1,  |,  ^, 
&c.?  Ans.   VVoW- 

18.  A  man  purchased  a  house,  giving  1  dollar  for  the  first 
door,  2  dollars  for  the  second,  4  dollars  for  the  third,  and  so 
on,  there  being  1 0  doors.     What  did  the  house  cost  him  ? 

Ans.  ^1023. 

(Art.  113.)  By  equation  (2),  and  the  rule  subsequently 

given,  we  perceive  that  the  sum  of  a  series  depends  on  the 

first  and  last  terms  and  the  ratio,  and  not  on  the  number  of 

terms;  and  whether  the  terms  be  many  or  few,  there  is  no 


GEOMETRICAL  PROGRESSION.  217 

variation  in  the  rule.  Hence,  we  may  require  the  sum  of  any 
descending  series,  as  1,  ^,  j,  |,  &c.,  to  infinity,  provided  we 
determine  the  last  term.  Now,  we  perceive  the  magnitude  of 
the  terms  decrease  as  the  series  advances;  the  hundredth 
term  would  be  extremely  small,  the  thousandth  term  would 
be  very  much  less,  and  the  infinite  term  nothing ;  not  too  small 
to  be  noted,  as  some  tell  us,  but  absolutely  vMhing. 

Hence,  in  any  decreasing  series,  when  the  number  of  terms 
is  conceived  to  be  infinite,  the  last  term,  L,  becomes  0,  and 
equation  (2)  becomes 

— a 

By  change  of  signs   .     s=- 

This  gives  the  following  nile  for  the  sum  of  a  decreasing 
infinite  series : 

Rule  . — Divide  the  first  term  hy  the  difference  hetvxen  unity 
and  the  ratio. 

EXAMPLES. 

1.  Find  the  value  of  1,  f ,  t\,  &c.,  to  infinity. 

a=r\f  r=f  Ans.  4. 

/  2.  Find  the  exact  value  of  the  series  2,  1,  ^,  &c.,  to  infinity. 

/  Ans.  4. 

t    3.  Find  the  exact  value  of  the  series  6,  4,   &c.,  to  infinity. 

Ans.  18. 

4.   Find  the  exact  value  of  the  decunal  .3333,  &c.,  to 

infinity.  Ans.    \. 

This  may  be  expressed  thus :  fV+rf  7'  <^c-    Hence,  a=^, 

6.  Find  the  value  of  .323232,  <fec.,  to  infinity.  m 

«=t¥5  »  «»'=«T5¥5iri  therefore,  r«xiv         -4nir.(|f. 
19 


218  ELEMENTARY   ALGEBRA,   r 

6.  Find  the  value  of  .777,  &c.,  to  infinity.    .     .     Am.     |. 

7.  Find  the  sum  of  the  infinite  series  1  +  -^H — ^H — 6+,  &c. 

Ans.  - — r- 
x^ — 1 

8.  Find  the  sum  of  the  infinite  geometrical  progression 

a — h-\ -A — ^ — ,  (fee,  in  which  the  ratio  is . 

a       a^     a?  a  ^2 

Am.  —r- 
a-\-o 

(Art.  114.)  When  three  numbers  are  in  geometrical  pro- 
gression, the  product  of  the  extremes  is  equal  to  the  square 
of  the  mean. 

This  principle  is. obvious  from  the  general  series 
a,  ar,  ar^,  ar^,  ar*,  ar^,  &c. 

Taking  any  three  consecutive  terms  anywhere  along  the 
series,  we  observe,  that  the  product  of  the  extremes  is  equal  to 
the  square  of  the  mean. 

That  is,  if  the  three  terms  taken,  are  a,  ar,  ar^, 
a^r^=(ary 

If  ar^,  ar^,  ar"^  are  the  three  terms, 
a7^Xa7^=(a7^y 
Hence,  to  find  a  geometrical  mean  between  two  numbers,  we 
must  multiply  them  together,  and  take  the  square  root.     ^ 
we  take  four  consecutive  terms,  the  product  of  the  extremes  will 
he  equal  to  the  product  of  the  means. 

(Art.  115.)  This  last  property  belongs  equally  to  geome- 
trical proportion,  as  well  a^  to  a  geometrical  series,  and  the 
learner  must  be  careful  not  to  confound  proportion  with  a 
series. 

a: ar:  :h:hr,  is  a  geometrical  proportion,  not  a  continued 
series.  The  ratio  is  the  same  in  the  two  couplets,  but  the 
magnitudes,  a  and  h,  to  which  the  ratio  is  applied,  may  be 
very  diflferent. 


GEOMETRICAL  PROGRESSION.  219 

We  may  suppose  a  :  ar  two  consecutive  terms  of  one  series, 
and  b :  br  any  two  consecutive  terms  of  another  series  having 
the  same  ratio  as  the  first  series,  and,  being  brought  together, 
they  form  a  geometrical  proportion.  Hence,  the  equality  of  the 
ratio  constitutes  proportion. 

EXAMPLES. 

1.  Find  the  geometrical  mean  between  2  and  8.    Ans.     4. 


(Art.  114)         ^2X8=4 

2.  Find  the  geometrical  mean  between  3  and  12. 

Ans.     6. 

3.  Find  the  geometrical  mean  between  5  and  80. 

Ans.  20. 

4.  Find  the  geometrical  mean  between  a  and  b. 

Ans.  {aby, 

5.  Find  the  geometrical  mean  between  \  and  9.    Ans.    |, 

6.  Find  the  geometrical  mean  between  3a  and  27a. 

Ans.  9a. 

7.  Find  the  geometrical  mean  between  1  and  9.    Ans.    3. 

8.  Find  the  geometrical  mean  between  2  and  3.  Ans.  JQ, 

9.  Find  two  geometrical  means  between  4  and  256. 

N.  B.  When  the  two  means  are  found,  the  series  will  con- 
sist oi  four  terms,  and  4  will  be  the  first  term  and  256  will  be 
the  last  term. 

Comparing  this  with  the  general  series, 

a,  ar,  ar^,  aT^y  we  have 
a=:4  and  ar'=256 
Hence,     .     .     .    r^=64    or    r=4 
Therefore,  16  and  64  are  the  means  required. 

10.  Find  three  geometrical  means  between  1  and  16. 


220  ELEMENTARY   ALGEBRA. 

Here,  the  first  term  of  tlie  series  is  1,  the  last  term  IG,  and 
the  number  of  terms  5,  because  three  terms  are  required,  and 
two  are  ah'eady  given. 

Now,  by  equation  (1),     L^^ar'^"^ 

That  is,    ....     ar"-^=16 

But  as  a=l,  and  »=5,  this  equation  is 

Hence,    .     .     •    .     •     r=2 
Therefore,  the  means  required  are  2,  4,  and  8. 
We  may  obtain  the  ratio  when  the  first  and  last  terms  are 
given,  by  the  following  formula  :         ^ 


\  a 


1 1 .  The  first  and  last  terms  of  a  geometrical  series  are 
2  and  162,  and  the  number  of  terms  5  ;  required  the  ratio. 

Ans.  3 

12.  The  first  term  of  a  geometrical  series  is  28,  the  last 
term  17500,  and  the  number  of  terms  6;  what  is  the  ratio? 

Ans.  5. 

PROBLEMS     THAT     INVOLVE     THE     PRINCI- 
PLES   OF    GEOMETRICAL    PROGRESSION. 

(Art.  116.)  When  we  wish  to  express  three  unknown 
quantities  in  geometrical  progression,  we  may  represent  them 
by  X,  Jxy,  y,  or  by  x^y  ry,  y^,  or  by  Xy  icy,  xy'^y  for  either  of 
these  correspond  with  Art.  114;  that  is,  the  product  of  the 
extremes  is  equal  to  the  square  of  the  mean. 

When  we  wish  to  express  four  unknown  quantities  in  geo- 
metrical progression,  we  may  express  them  by  ar,  ary,  ay*,  ary*, 
or  by  P,  X,  y,  Q. 

The  object  of  this  last  notation,  is  to  reduce  P  and  Q  to 
terms  expressed  by  x  and  y,  thus : 


GEOMETRICAL  PROGRESSION.  221 

Taking  the  first  three  terms  only,  we  shall  have 

Or, P=-        .     ,     . 

y 

Taking  the  last  three  terms  only,  we  shall  have 

Therefore,  four  quantities  in  geometrical  progression  may 
be  expressed  by  x  and  y  only^  and  the  terms  stand  symmetri- 
cally thus  : 

x"  y" 

7.'  ^'  y^  t: 

y  X 

In  a  similar  manner,  we  might  express  more  terms  by  x 
and  y  only,  and  have  them  stand  symmetrically,  if  it  were 
proper  to  extend  this  subject  in  a  work  as  elementary  as  this. 

1.  Three  numbers  are  in  geometrical  progression,  the  sum 
of  the  first  and  second  is  90,  and  the  sum  of  the  second  and 
third  is  1 80.     What  are  the  numbers  ? 

Am.  30,  60,  and  120. 
Represent  the  numbers  by  a;,  xy,  and  xy^. 

2.  The  sum  of  three  numbers  in  geometrical  progression 
is  7,  and  the  sum  of  their  squares  is  21.  "What  are  the 
numbers  ?  Ans.  1 ,  2,  4. 

This  problem  furnishes  the  following  equations : 

a:+»Jx2j-\ry=7  (1) 

x'+xy+y'^n  (2) 

From  (1)       ,     .     .     .     X'\-y—a—J^     (3) 

From  (2)       ....  ^-\-f^^a—xy      (4) 

Squaring  (3),     .  x^-{-^xy-\-y'^=c?-^'iLaJxy\-xy  (5) 

Subtracting  (4)  from  (5),  ^xy=a^-^Za—^aJxy-\-2xy  (6) 


222  ELEMENTARY  ALGEBRA. 

Dropping  9.xy  from  botli  members,  dividing  by  a,  and 
transposing,  we  have  .     .   ^Jxy=^a — 3 

That  is, V^=2         (7) 

This  value  of  Jxy  put  in  equation  (3),  gives 

3.4.^=5        (8) 

From  equations  (7)  and  (8),  we  find  x  and  y,  as  taught  in 
Art.  101. 

3.  The  sum  of  the  first  and  third  of  four  numbers  in  geo- 
metrical progression  is  20,  and  the  sum  of  the  second  and 
fourth  is  60.     What  are  the  numbers  ?        Ans.  2,  6,  1 8,  54. 

4.  Divide  the  number  210  into  three  parts,  so  that  the  last 
shall  exceed  the  first  by  90,  and  the. parts  be  in  geometrical 
progression.  Ans.  30,  60,  and  120. 

5.  The  sum  of  four  numbers  in  geometrical  progression  is 
30;  and  the  last  term  divided  by  the  sum  of  the  mean  terms 
is  li.     What  are  the  numbers?  Ans.  2,  4,  8,  and  16. 

6.  The  sum  of  the  first  and  third  of  four  numbers  in  geo- 
metrical progression  is  148,  and  the  sum  of  the  second  and 
fourth  is  888.     What  are  the  numbers  ? 

Ans.  4,  24,  144,  and  864. 

7.  The  continued  product  of  three  numbers  in  geometrical 
progression  is  216,  and  the  sum  of  the  squares  of  the  ex- 
tremes is  328.  What  are  the  numbers  ?  Ans.  2,  6,  18. 

8.  The  sum  of  three  numbers  in  geometrical  progression 
is  13,  and  the  sum  of  the  extremes  being  multiplied  by  the 
mean,  the  product  is  30.     What  are  the  numbers  ? 

Ans.  1,  3,  and  9. 

9.  There  are  three  numbers  in  geometrical  progression 
whose  product  is  64,  and  the  sum  of  their  cubes  is  584. 
What  are  the  numbers  ?  Ans.  2,  4,  and  8. 


GEOMETRICAL  PROGRESSION.  223 

Let  ar^,  xy,  y^  represent  the  three  numbers. 

Then, x'if^Q\  (1) 

(^■\-j^y^-\-y^=bZ^  (2) 

.     .   ar^^  .  .  =64 


Also,  .  . 
Add,  .  . 
And  .  . 
Square  root 


a:«4-2a^2/2+2/^=648=324*2  (3) 
.     .       ^■^f=\zji  (4) 

From  (2)  subtract  three  times  (1),  and  we  have 
JB«— 2a^/-f2^=392=196'2  (5) 

Square  root,  .     .     .       y^ — x^=\^J^  (6) 

We  give  the  minus  sign  to  s^,  because  y  must  be  greater 
than  X  from  the  position  it  occupies  in  our  notation,  and  o^ — y^ 
or  y^ — 01^ y  when  squared,  will  produce  the  same  power. 

Subtracting  (6)  from  (4),  and 

2ar^=4<y2 

Or, Q?=2j^ 

Squaring,     ....       a;^=8 

Cube  root,   ....      ar*=2,     Ans. 

10.  There  are  three  numbers  in  geometrical  progression, 
the  sum  of  the  first  and  last  is  52,  and  the  square  of  the 
mean  is  100.     What  are  the  numbers?  Ans.  2,  10,  50. 

11.  There  are  three  numbers  in  geometrical  progression, 
their  sum  is  31,  and  the  sum  of  the  squares  of  the  first  and 
last  is  626.     What  are  the  numbers  ?  Ans.  1,  5,  25. 

12.  It  is  required  to  find  three  numbers  in  geometrical  pro- 
gression, such  that  their  sum  shall  be  14,  and  the  sum  of 
their  squares  84.  Ans.  2,  4,  and  8. 

13.  There  are  four  numbers  in  geometrical  progression,  the 
second  of  which  is  less  than  the  fourth  by  24;  and  the  sum 
of  the  extremes  is  to  the  sum  of  the  means,  as  7  to  3.  What 
are- the  numbers  ?  An^.  1,  3,  9,  and  27. 


224  ELEMENTARY  ALGEBRA. 

14.  The  sum  of  four  numbers  in  geometrical  progression 
is  equal  to  the  common  ratio  +1,  and  the  first  term  is  j\. 
What  are  the  numbers  ?  Ans.  j\,  j\,  j%,  3.7. 


PROPOETION. 

(Art.  117.)  Two  magnitudes  of  the  same  kind  can  be 
compared  with  each  other,  and  the  numerical  relation 
between  them  determined.  The  manner  of  determining  this 
relation,  is  to  divide  one  by  the  other,  and  the  quotient  is 
called  the  ratio  betweenthe  two  magnitudes.  When  two  quan- 
tities have  the  same  ratio  as  two  other  quantities,  the  four 
quantities  may  constitute  a  proportion. 

Therefore,  proportion  is  the  equality  of  ratios. 

Proportion  is  written  in  two  ways, 

Thus, aib:  :c:d 

Or  thus,      .     .     .     .     a:h  =  c:d 

The  last  is  the  modern  method,  and  means  that  the  ratio 
of  a  to  5  is  equal  to  the  ratio  of  c  to  d. 

If  a  is  taken  as  the  unit  of  measure  between  a  and  h,  then 

-  IS  the  numerical  ratio  between  these  two  magnitudes. 

If  c  is  taken  for  the  unit  of  measure  between  c  and  d,  then 

d. 

-  is  the  numerical  ratio  between  these  two  magnitudes. 
c  ^ 

The  magnitudes  a  and  h  may  be  very  different  in  kind  from 
those  of  c  and  d;  for  instance,  a  and  b  may  be  bushels  of 
wheat,  and  c  and  d  sums  of  money. 

This  manner  of  comparing  magnitudes,  by  taking  one  of 
them    as   a  whole    (regardless   of  other   units)   is   called 


PROPORTION.  225 

geometrical  proportion,  and  if  there  are  more  than  two  magni- 
tudes having  the  same  ratio,  the  magnitudes  are  said  to  be  in 
geometrical  progression. 

Two  magnitudes  compared  by  raiio  are  called  a  couplet. 
Thus,  a  :  5  is  a  couplet,  and  c.dis  another  couplet. 

The  first  magnitude  of  a  couplet  is  called  the  antecedent^ 
the  second  the  conseqtteni. 

A  ratio  can  exist  between  two  magnitudes ;  but  a  proportion 
requires  four — tivo  antecedents  and  two  consequents  having  the 
same  ratio. 

Thus,  if    .     .     ,     .    a:  h=c :  d 

Then -=-  by  the  def.  of  proportion, 

0)      c 

All  operations  in  proportion  rest  on  this  fundamental  equa- 
tion ;  and  to  prove  a  principle  or  an  operation  true,  we  directly, 
or  remotely  compare  the  principle  or  the  operation  to  this 
equation,  and  if  we  find  a  correspondence,  the  principle  or 
the  operation  is  true — otherwise,  false. 

PROPOSITION    I. 

In  every  proportion,  the  product  of  the  extremes  is  equal  to  the 
product  of  the  means. 

Let a: b=c  : d  represent  any  proportion, 

b     d 

Then, -=-  must  be  a  true  equation. 

a     c  ^ 

Multiply  both  members  of  this  equation  by  ac,  and  as  the 

product  of  equal  factors  are  equal  (Ax.  3), 

Therefore,    .     .     .     .     cb=ad 

That  is,  the  product  of  c  and  b,  the  means,  is  equal  to  the 
product  of  a  and  d,  the  extremes. 

S  c  H  o  L I  u  M. — Divide  both  members  of  this  equation  by  a, 

Then -=ti 

a 


226  ELEMENTARY  ALGEBRA. 

This  equation  shows,  that  the  fourth  term  of  any  proportion 
may  be  found  from  the  first  three,  by  the  following 

Rule  . — Multiply  the  second  and  third  terms  of  the  proper' 
tion  together,  and  divide  that  product  hy  the  first  term. 

This  is  a  part  of  the  well  known  rule  of  three,  in  Arithmetic. 

PROPOSITION    II. 

Conversely.  ]f  the  product  of  two  quantities  is  equal  to  the 
product  of  two  others,  then  two  of  them  may  he  taken  for  the 
means,  and  the  other  two  for  the  extremes  of  a  proportion. 

Let ch=ad 

Divide  both  members  of  this  equation  by  any  one  of  the 
four  factors,  say  c,  then  we  have 

ad 
c 
Divide  this  last  equation  by  another  of  the  factors,  say  a. 

Then -=- 

a     c 

This  is  the  fundamental  equation  for  proportion,  and  gives 
a :  b=c  :d 

Now,  as  the  principle  is  established,  we  may  proceed  more 
summarily,  and  take  the  two  factors  in  one  member  for  the 
extremes, 

Thus,      ....     a:      =      :d 

To  fill  up  the  means,  we  must  take  the  factor  which  has 
the  same  name  as  a  to  stand  before  the  equality,  and  the 
other  factor  to  stand  after  the  equahty  will  be  of  the  same 
name  as  d,  and  the  proportion  will  be  complete. 

If  the  quantities  are  all  numerals,  it  is  immaterial  which 
factor  stands  first  in  the  means. 

Thus,  a :  b=c  :  d  )   are  proportions  equally  true  in  numeri- 

Or,  .   a :  c=6  :d  )   cal  values. 


PROPORTION.  227 

Scholium . — A  proportion  and  an  equation  may  be  re- 
garded as  but  a  different  form  for  the  same  expression,  and 
every  equation  may  be  put  into  a  proportion.     For  example, 

Wliat  proportion  is  equivalent  to  the  following  equation  ? 
(xy=a{a-\-b) 

Ans.      .     .     .     ;      a;:a=a-l-J:y, 

Or, (cy:  a=^a-{-h :  1 

Or, a  :  rr=y :  {a-\-h) 

What  proportion  is  equivalent  to  the  equation 


cd 
a 
Thus,  we  might  give  examples  without  end. 


Ans,     .     ,  x\  {  — |- 1  j  =a :  1 


PROPOSITION    III. 

If  three  quantities  are  in  continued  proportion,  the  product 
of  tlie  extremes  is  equal  to  the  square  of  the  mean. 

If a:  b=b  :  c 

From  proposition  1,      ac=bb=h'^ 
That  is,  if  proposition  1  is  true,  the  truth  of  this  proposition 
follows  as  an  inevitable  consequence. 

If  ac=b^,  then  b=Jac,  which  shows  that 

The  mean  proportional  between  two  quantities  is  found  by 
extracting  the  square  root  of  their  product. 

PROPOSITION    IV. 

If  four  quantities  are  in  proportion,  they  will  be  in  proportion 
by  INVERSION,  that  is,  the  second  will  be  to  the  first,  as  the  fourth 
to  the  third. 

Let a'.b=:^c:d 


228  ELEMENTARY  ALGEBRA. 

Then,  by  the  definition  of  ratio  and  proportion,  we  have 

a     c 

h  .    a 

Divide  1  by  -  and  the  quotient  is  -. 

d  .    c 

Divide  1  by  -  and  the  quotient  is  -. 

But  equals  divided  by  equals  must  produce  equal  quotient* 
(Ax.  4). 


Therefore,     .     . 

a     c 

Or,      .... 

.     b:a=  die 

In  numbers  if     . 

.     3:5=12:20 

Then    .... 

.     5:3=20:12 

PROPOSITION     V. 

Magnitudes  which  are  proportional  to  the  same  proportionals, 
are  proportional  to^each  other. 

If      a:b=F:Q 


a:b=F:  Q  ) 
c:d=P:Q  J 


.    ,        ,     ^    ^  ,       Then  a :  b=c :  d. 
And 

b     Q 

From  the  first  proportion,  -=p 

a      Jr 

By  the  second,     .     .     .     -=5 

C      Jr 

Therefore,       ....    -=-     (Ax.  1). 
a     c     ^ 

Or, a\b=^c\d 


PROPOSITION     VI. 

If  four  magnitudes  be  in  proportion^  they  must  be  in  propor- 
tion by  COMPOSITION ;  that  is,  the  first  will  be  to  the  sum  of  the 
first  and  second,  as  the  third  will  be  to  the  sum  of  the  third  and 
fourth ;  and  the  first  is  to  the  difference  between  the  first  and 
second,  as  the  third  is  to  the  difference  between  the  third  and  fourth. 


PROPORTION.  229 

On  the  supposition  that  .  a :  5=c :  d 
We  are  to  prove  that  a :  a-\-h=^c :  c-j-rf 

T.  ,  .  .  h    d 

From  the  supposition,     .       -=- 

Add  each  member  of  this  equation  to  unity,  and  then  we 

have 1+-=1+- 

a  c 

Reducing  these  mixed  quantities  to  improper  fractions, 

Ajid a+6^c+d 

a  c 

That  is,  .     .     .     .     a:  a-|-5=c :  c-\-d 

Subtracting  each  member  of  the  original  equation  from 
unity,  and  we  have 

a  c 

a — b    c — d 


ur,    .     .     •     •     • 

.    = 

a          c 

Therefore,  .     .     . 

a :  a — 5=c :  c — d 

Scholium . — This  composition  may  be  carried  to  almost 

ly  extent,  as  we  see 

by  the  following  investigation : 

Take  Ae  equation. 

b_d 

a     c 

Multiply  both  members  by  m,  then 

mb^md 
a       c 

Add  each  member  of  this  equation  to  n,  ,. 

Then,      .... 

,  nib        ,  md 
.    w+  — =wi — 
a             c 

By  reduction,    .     . 

na-\-mb    iic-{-md 
a               c 

Hence,    ...  a 

•.na'\-mb=sQ'.nG-^md 

230  ELEMENTARY  ALGEBRA. 

PROPOSITION     VII. 

If  four  quantities  he  in  proportion,  the  sum  of  the  two  quan- 
iities  which  form  the  first  couplet  is  ip  their  difference,  as  the  sum 
of  the  two  quantities  which  form  the  second  couplet  is  to  their 
difference. 

On  the  supposition  that  •     •    a  :  5=c :  d 

We  are  to  prove  that     a-\-h :  a — 5=c+a? ;  c — d 

From  proposition  6,  .     .     a\a-\-h^=c\c-\-d        (1) 

Also, a\a — h=c\c^d        (2) 

■n         /H\  a-\-h    c-\-d 

From(l), __L-=-_L_ 

^  '  a  c 

Dividing  both  members  of  this  equation  by  {a-\-h),  and 
multiplying  both  members  by  c,  we  have 

c c-\-d 

a    a+d 

Operating  in  the  same  manner  with  (2),  we  shall  find 

c    c — d 
a     a — b 


c-\-d    c-    w      -  .       . . 
Therefore, — r7= 1     (Ax.  1). 

a+o     a — 0     ^  ^ 

Whence,     ....    a-]rh\c-\-d=za — b:c — d 
Or, a — h  :  c — d=a-]rh :  c-{-d 

PROPOSITION     VIII. 

If  four  quantities  he  in  proportion,  either  couplet  may  he  mul- 
tiplied or  divided  hy  any  number  whatever,  and  the  quantities 
will  still  he  in  proportion. 

Let   ........    a :  h=c :  d 

Then, -=- 


PROPORTION.  .        231 

Multiplying  both  numerator  and  denominator  of  either  of 
these  fractions  by  any  number,  n. 

Then, — =- 

na     c 

b     7id 

Also, -= — 

a     nc 

That  is,    ,     .     .     ,    na:  nb=^c :  d 

Also,        a\  h=nc :  nd 

Here,  n  may  represent  any  number  whatever ;  and  if  it 
represents  a  whole  number,  as  3,  7,  8,  &c.,  then  the  couplet 
is  multiplied.  If  n  represent  a  fraction,  as  i,  i,  |,  &c.,  then 
the  couplet  is  divided. 

PROPOSITION    IX. 

If  four  quantities  be  in  proportion,  the  antecedents  may  be 
multiplied  by  any  numJyer,  and  they  will  still  be  in  proportion; 
also,  the  consequents  may  be  multiplied  by  any  number,  and  the 
four  quantities  will  still  be  in  proportion. 

Let a:  b=.c :  d 

Then .    *=^ 

a     c 

Multiplying  this  equation  by  m,  then 

mb md 

a       c 

Therefore,    . '   .     .     a: 7rib=c  -.md 

Divide  both  members  of  the  original  equation  by  m. 

Then, ±=-^ 

ma    mc 

Hence,    .     .    .     .     ma:  b=m^ :  d 


232  ELEMENTARY  ALGEBRA, 

PROPOSITION    X. 

If  four  magnitudes  he  in  2}roportionf  like  powers  or  roots  of 
the  same  will  be  in  proportion. 

Let     ......    a:h=c'.d 

Then, *=^ 

a     c 

Raise  both  members  of  this  equation  to  any  power  denoted 
by  n. 

Then, — =— 

Hence,"  ....       a»:5"=c":c?» 
By  extracting  any  root  of  the  primitive  equations,  which 
may  be  designated  by  -,  we  have 


1 
1 "~" 

1 

a« 

c" 

1 

1 

I      1 
:c^ :  (f 

Hence,      .    .    . 

PROPOSITION    XI. 

If  four  quantities  he  in  proportion^  also  four  others,  the 
^product  or  quotient  of  the  two,  term  ly  term,  will  still  form 
proportions. 

If a:h=c'.d 

And, x'.y=^m:n 

Then  we  are  to  prove  that 

ax :  hy=cm :  dn 

.    .  abed 

And, -:-=-  :- 

X  y    m   n 

From  the  first  proportion  we  have 

U.     (,) 

a     c         ^   ^ 

From  the  second,    .     .    -=-       (2) 
X    m       ^  ^ 


PROPORTION.  233 

By  multiplying  these  two  equations  together,  term  by  term, 
ire  find 

ly nd 

ax    mc 
That  is,  .     .     .     .     ax: bi/=mc : nd 
Apply  proposition  1  to  the  two  given  proportions,  and  we 

have ad=bc 

And, nx=.my 

Dividing  one  of  these  equations  by  the  other,  we  have 

\x)  Kni       \y)\m) 
By  the  reverse  application  of  proposition  1 ,  we  have 
ah     c    d 
X  '  y    m  '  n 

PROPOSITION    XII. 

If  any  number  of  proportionals  have  the  same  ratio,  any  one 
vf  the  antecedents  will  be  its  consequent,  and  as  the  sum  of  all  the 
antecedents  to  the  sum  of  all  the  consequeMs. 

Let a  :  b=  a  :  h 

Also,    .     .     .     .     .     a :  5=  c :  d 
a:b=m:n 
(fec.=&c. 
Then  we  are  to  prove  that 

a:b=(a-\-c-\-m):(b-^d-\-n) 
From  the  first  prop.,       ab=ab 
From  the  second,      .     ad=cb 
From  the  third,    .     .     an=nb 
By  addition,     a(b-{-d-\-n)=b(a-\-c-]-m) 
By  prop.  1,  .     .     .     a:b=(a-\rc-^m):(b-{-d-{-n) 

The  following  examples  are  intended  to  illustrate  the  prac- 
tical utility  of  the  foregoing  propositions : 
20 


234 


ELEMENTARY  ALGEBRA 


EXAMPLES. 

1.  Find  two  numbers,  the  greater  of  which  is  to  the  less  as 
their  sum  to  42,  and  the  greater  to  the  less  as  their  difference 
is  to  6. 

Let   .    x=  the  greater  and  y=  the  less. 

mi        1  Tx-  (  x:y=x-\-y:42 

Then,  by  conditions      i  „ 

•'  (  x: y=-x — y : 6 

(Prop.  6),       .     .    ic+y:42=a; — y\^ 

Changing  means,  x-\-y :  x — ?/=42  : 6 

(Prop.  9),      .     .     .    2a;:2y=48:36 

(Prop.  8),       .     .     .      a; :  y=  4  : 3 

With  these  proportions  of  x  and  y,  we  return  to  the  original 

conditions ;  applying  proposition  5,  and  we  have 

4  :  ^—x-\-y  :  42 

4  :  3=a;— y  :  6 
From  the  first,    .     .      a;-}-y=56 
From  the  second,      .      x — ^2/=  8 
Hence, ic=32,  y=24 

2.  Divide  the  number  14  into  two  such  parts,  that  the  quo- 
tient of  the  greater,  divided  by  the  less,  shall  be  to  the  less, 
divided  by  the  greater,  as  100  to  16. 

Let  .  a:=  the  greater,  and  y=  the  less  part, 

Then, -:^=100:16 

y   X 

And, a;+2/=14 

(Prop.    8),      ...  a^:y2=:i00:16 

(Prop.  10),      .     .     .     ar:y=10:4 

Hence, 9,x=5y 

But, x-{-y=14 

Therefore, a:=10,  y=4 

3.  Find  three  numbers  in  geometrical  progression  whose 
sum  is  13,  and  the  sum  of  the  extremes  is  to  the  double  of 
the  mean  as  10  to  6. 

Let  a:,  xy,  and  xy^  represent  the  numbers, 


PROPORTION. 

Then,  by  the  conditions,        x-\rX2/-\-xy^=\3 

And, xf-\-x :  2xy=10  :  e 

(Prop.    8), 2/2+l:2y=10:6 

(Prop.    7),  (y2+2y+l):(y'— 2y+l)=16:4 

(Prop.  10), y+l:y— 1=4:2 

(Prop.    8), 2y:2=6:2 

Hence, y=3,  x=l 

4.  The  product  of  two  numbers  is  35,  and  the  difference 
of  their  cubes  is  to  the  cube  of  their  difference  as  109  to  4. 
What  are  the  numbers  ?  Ans.  7  and  5. 

Let  X  and  y  represent  the  numbers.  Then,  by  the  given 
conditions, xy=^35 

And, a^—y'^:(x—yy=\09:4 

Divide  the  first  couplet  by  (x — y)  (Prop.  8).  Then  we 
have       ....     x^-{-xy-\-f :  (x—i/y=\09  :  4 

Expanding  (x — yy,  and  then  making  appHcation  of  (Prop. 
7),  we  have     ....       Sxy:(x — yY=105:4 

But,         3ary=105 

Therefore, (x — yy=4: 

And, .     X — y=2 

5.  What  two  numbers  are  those,  whose  difference  is  to 
their  sum  as  2  to  9,  and  whose  sum  is  to  their  product  as  1 8 
to  77?  Am.  11  and  7. 

6.  Two  numbers  have  such  a  relation  to  each  other,  that 
if  4  be  added  to  each,  they  will  be  in  proportion  as  3  to  4; 
and  if  4  be  subtracted  from  each,  they  will  be  to  each  other 
as  1  to  4.     What  are  the  numbers  ?  Ans.  5  and  8. 

7.  Divide  the  number  16  into  two  such  parts  that  their 
product  shall  be  to  the  sum  of  their  squares  as  15  to  34. 

Ans.  10  and  6, 

8.  There  are  two  numbers  whose  product  is  320;  and  the 
difference  of  their  cubes,  is  to  the  cube  of  their  difference,  as 
61  to  1.     What  are  the  numbers?  Ans.  20  and  16. 


236  ELEMENTARY  ALGEBRA. 

CONCLUSION. 

We  conclude  this  volume  by  giving  a  general  investigation 
of  the  rules  in  Arithmetic  for  the  computation  of  interest,  and 
the  adjustment  of  accounts  in  fellowship. 

Interest  is  a  percentage  paid  for  the  tise  of  money  for  a  spe- 
cified time. 

On  this  single  definition,  all  the  rules  of  computation  are 
founded.     The  unit  for  time  is  commonly  one  year. 

Let  r  represent  the  interest  corresponding  to  unity  of  prin- 
cipal remaining  at  interest  for  unity  of  time. 

Then,  as  a  double  capital  would  demand  double  interest 
for  the  same  time,  a  treble  capital,  treble  interest,  and  so  on. 
Therefore,  if  P  represents  any  principal  or  capital,  we  have 
the  following  proportion : 

Trin.  Int.  Prin.    Int. 

1  :  r=F  :  rP 

The  last  term  of  this  proportion  shows  that  to  find  the  in- 
terest of  any  principal  for  one  year,  we  must  multiply  that 
principal  by  the  decimal  rate  per  cent.* 

For  double  the  length  of  time,  the  interest  must  be  double, 
for  treble  the  length  of  time,  the  interest  must  be  treble  ;  and 
so  on. 

Now,  let  t  represent  the  length  of  time  that  any  principal, 
P,  remains  at  interest,  r  being  the  rate  per  cent.,  and  /  the 
aggregate  interest,  then  we  shall  have  this  general  equation. 
Prt=I        (1) 

This  gives  the  following  universal  rule  for  computing 
interest. 

Rule  . — Multij^ly  the  principal  hy  the  decimal  rate  per  cent., 
and  that  product  by  tltc  time, 

*  Rate  per  cent  is  but  another  definition  for  the  interest  of  unity  of  prin- 
cipal for  unity  of  time. 


PROPORTION.  237 

If  we  consider  that  the  principal  and  interest  added  together 
must  give  the  amount,  and  if  we  put  A  to  represent  the 
amount,  then  we  shall  have 

Fri-hF=A        (2) 

Equations  (1)  and  (2)  embrace  all  the  conditions  in  relation 
to  interest,  and  furnish  all  the  rules  for  computations. 
For  instance,  equation  (1)  gives 

Equation  (2)  gives    .     P=   j— 

That  is,  when  any  problem  requires  the  finding  of  the 
principal,  observe  the  following  rules  : 

R  u  L  E  1 . — Divide  the  interest  ly  the  ^product  of  the  rate  and 
time. 

R  u  L  E  2 . — Divide  the  amount  ly  the  product  of  the  rate 
and  tivMy  increased  ly  unity. 

Equation  ( 1 )  gives    .      ^—-p 

That  is,  to  find  the  time,  we  have  the  following  rule  : 

Rule  . — Divide  the  whole  interest  hy  the  interest  for  one 
year. 

Equation  (1)  gives     .      ^='p> 

To  find  the  rate  per  cent.,  take  the  following  rule : 
Rule  . — Divide  the  interest  hy  the  product  of  the  principal 
and  time. 

FELLOWSHIP. 

Two  men  united  capital  to  engage  in  a  certain  enterprise, 
the  first  put  in  a  dollars,  the  second  I  dollars,  and  they  gained 
g  dollars.     Give  a  rule  for  the  equitable  division  of  this  gain. 


ELEMENTARY   ALGEBRA. 

Let  X  represent  the  portion  belonging  to  tliat  one  which  paid 
in  a  dollars,  and  y  the  portion  of  the  other. 

Then,     ....     x+y=ff 

But  their  portions  of  the  gain  should  be  in  just  the  same 
proportion  as  their  capital  paid  in, 

That  is,  ...     .      a? :  y=a  :  b,  or  hx=ay 

Multiply  the  first  equation  by  a,  then 

ax-\-ay=agy  or  ax-{-hx=ag 

Or, .=^,    y^  ^ 


'a-\-b*     ^     a^b 

Hence,  we  have  the  following  rule  to  find  each  man's  share. 

Rule  . — Multiply  the  gain  by  each  man's  stock,  and  divide 
the  product  by  the  whole  capital  invested. 

Again,  suppose  three  persons.  A,  B,  and  C,  enter  into 
partnership,  and  furnish  capital  in  proportion  to  a,  b,  and  c, 
and  they  gain  a  sum,  g,  what  is  each  man's  share  of  it  ? 

Let     .     x=  A's  share,  y=  B's  share,  and  ^=C's  share, 


Then,      .     .     .   x-jry-^z=g 

(1) 

And, x:y=a:b, 

also, 

y:z=a:c 

From  the  first  propor.,   y=- 

(2) 

ex 
From  the  second,  .     .     «=  — 

a 

(3) 

These  values  of  y  and  2,  put  in  equation  (1),  give 

bx    ex 

x-i h— =0',  or  ax-\-ox-]rcx=:.ag 

a     a 

Or ^=-4^^        (4) 

a-\-b-\-c         ^   ' 

This  value  of  x  put  in  (2)  gives  y,  and  in  (3)  gives  z. 


FELLOWSHIP.  239 

Here,  again,  we  find  that  each  man's  share  of  the  gain  is 
equal  to  the  whole  gain  multiplied  hy  his  particular  portion  oj 
Ike  stock,  and  that  product  divided  hy  the  whole  stock  invested. 

The  same  results  would  be  obtained  in  relation  to  any  nuni- 
ber  of  partners.  Observe,  that  g  can  be  of  any  value,  posi- 
tive, negative,  or  zero.  When  it  is  zero,  each  numerator  is 
zero ;  and,  thus,  x,  y,  and  z  becopae  zero,  as  they  ought 
in  that  case.  When  g  is  negative,  it  denotes  loss,  and  losses 
must  be  shared  in  the  same  proportion. 

It  is  not  necessary  that  a,  h,  and  c  should  designate  the 
actual  stock  of  each  partner  if  they  represent  their  due  pro- 
portional parts. 

In  taking  up  a  book  on  common  Arithmetic,  we  find  the 
following  rule  for  fellowship  : 

As  the  whole  amount  of  stock  or  labor 

Is  to  each  Trumps  portion. 

So  is  the  whole  property,  loss  or  gain^ 

To  each  man's  share  of  it. 

These  four  lines  express  either  one  of  the  equations,  (4), 
(5),  or  (6);  for,  by  resolving  (4),  for  example,  into  a  propor- 
tion, we  have     .     .     (a-{-b-jrc)  :a=g:x 

Thus,  we  perceive  that  this,  like  most  other  arithmetical 
rules,  is  the  result  of  algebraic  investigation. 

Let  us  now  consider  the  case  in  which  time  is  an  element, 
and  for  the  sake  of  clearness  we  will  suppose  an  example. 

Two  men,  A  and  B,  hired  a  pasture,  for  which  they  agreed 
to  pay  g  dollars.  A  ptU  in  a  coivs  3  weeks,  B  put  in  b  cows 
for  6  weeks;  what  shall  each  pay  ? 

Consider  that,  a  cows  for  three  weeks  would  consume  as 
much  as  2>a  cows  for  one  week.  Also,  h  cows  for  five  weeks 
would  consume  as  much  as  bh  cows  for  one  week.  Thus,  we 
reduce  all  action  to  some  unit  of  time.  To  be  more  general,  we 
will  consider  3  and  5  as  t  and  t' ,  any  number  of  weeks  or 
days  whatever,  then  the  action  will  be  at  and  W,  and  it  is 


240  ELEMENTARY  ALGEBRA. 

evident  fhat  the  partners  must  pay  in  proportion  to  this  action, 
or  in  this  case,  to  the  consumption  of  the  cows. 

Now,  let  x=:  what  A  must  pay,  and  y=  what  B  must  pay. 

Then,     ....    a;-|-y=:^  (1) 

And,      ....      x:y—at:hi! 

Hence, y=(--Ja;      (2) 

This  value  of  y  put  in  equation  (1),  gives 

x-{—x=g,  or  {ai^bt:)x={at)g     (3) 

Cll 

Equation  (4)  will  furnish  the  following  proportion: 

[at-\-ht')  :  ai=g  :  a 
Equation  (5),  at-{-bt' :  ht'r=g  \y 

Taking  up  a  work  on  Arithmetic,  I  found  the  following  rule 
for  computing  results  in  compound  fellowship. 

Rule  . — Multiply  the  active  agents  hy  the  time  each  was  in 
action.     Then  by  proportion. 

As  the  sum  of  the  products 
'  Is  to  each  2mrticular  product, 

So  is  the  whole  gain  or  loss 
To  each  man's  share  of  it. 
Now,  it  is  evident  that  the  words  of  this  rule  were  dictated 
by  the  preceding  proportions. 

THE    END. 


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